{"id":10,"date":"2024-04-04T07:30:14","date_gmt":"2024-04-04T04:30:14","guid":{"rendered":"https:\/\/sisu.ut.ee\/huvitavkeemia\/26-lahuse-molaarse-kontsentratsiooni-arvutamine\/"},"modified":"2025-03-27T10:10:33","modified_gmt":"2025-03-27T08:10:33","slug":"26-lahuse-molaarse-kontsentratsiooni-arvutamine","status":"publish","type":"page","link":"https:\/\/sisu.ut.ee\/huvitavkeemia\/26-lahuse-molaarse-kontsentratsiooni-arvutamine\/","title":{"rendered":"2.6. Lahuse molaarse kontsentratsiooni arvutamine"},"content":{"rendered":"<p><span lang=\"et\"><span style=\"line-height: 115%;\">Nagu sa juba varasematest peat\u00fckkidest oled lugenud, on keemias \u00fcks p\u00f5hilisi suurusi ainehulk, mida kasutatakse erinevates arvutustes. Ainehulga kaudu saame reaktsiooniv\u00f5rrandite p\u00f5hjal arvutada, kui palju ainet reageerib v\u00f5i tekib. Samas tegeldakse keemias\u00a0 v\u00e4ga palju lahustega, mida saame m\u00f5\u00f5ta ruumala\u00fchikutes:\u00a0cm<sup>3<\/sup>, dm<sup>3<\/sup>, m<sup>3<\/sup>. Sellest tulenevalt ongi keemias \u00fcks peamisi kasutusel olevaid lahuse kontsentratsiooni v\u00e4ljendusviise\u00a0<\/span><\/span>molaarsus ehk molaarne kontsentratsioon. J\u00e4rgn<span lang=\"et\"><span style=\"line-height: 115%;\">evast videost saad vaadata, kuidas molaarset kontsentratsiooni arvutatakse, kuidas kasutatakse molaarset kontsentratsiooni ainehulga v\u00f5i siis lahuse ruumala leidmiseks ja reaktsiooniv\u00f5rrandiga arvutus\u00fclesannetes.<\/span><\/span><\/p>\n<p><\/p><div class=\"ratio ratio-16x9 mb-3\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--16x9\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n<h6>Allikas:\u00a0<a title=\"\" href=\"https:\/\/youtu.be\/Yykk-G7ww1A\" target=\"_blank\" rel=\"noopener\" data-url=\"https:\/\/youtu.be\/Yykk-G7ww1A\">https:\/\/youtu.be\/Yykk-G7ww1A<\/a><\/h6>\n<h5>Molaarsus\u00a0n\u00e4itab lahustunud aine hulka moolides \u00fches liitris lahuses.<\/h5>\n<p>Molaarsus\u00a0n\u00e4itab lahustunud aine hulka moolides \u00fches liitris lahuses ja\u00a0s<span lang=\"et\"><span style=\"line-height: 115%;\">eda t\u00e4histatakse t\u00e4hega <i>c<\/i><\/span><\/span>:<\/p>\n<p style=\"text-align: center;\">$c=\\frac{n}{V}$.<\/p>\n<p>V\u00f5rrandis on\u00a0<em>n<\/em>\u00a0ainehulk (\u00fchik mool, l\u00fchend mol) ja <em>V<\/em> on lahuse ruumala (\u00fchik dm<sup>3<\/sup>). <span lang=\"et\"><span style=\"line-height: 115%;\">Molaarsuse \u00fchikuks on mol\/dm<sup>3<\/sup> v\u00f5i l\u00fchendina M. J\u00e4ta meelde, et <i>V<\/i> on kogu lahuse ruumala.\u00a0<\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\">Kirjaviis 0,5 M lahus t\u00e4hendab, et 1 liitris (dm<sup>3<\/sup>) lahuses sisaldub 0,5 mooli lahustunud ainet. Molaarsuse arvutamiseks on vaja teada ainehulka <i>n<\/i> ja lahuse ruumala <i>V<\/i>. Toome \u00e4ra p\u00f5hilised valemid, mida v\u00f5ib veel vaja minna.<\/span><\/span><\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">Ainehulga arvutamise valemid:<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm; text-align: center;\">$n=\\frac{m}{M}$<\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\"><i>n\u00a0<\/i>\u2013 ainehulk\u00a0(\u00fchik mool, l\u00fchend mol)<\/span><\/span><\/span><br><span style=\"line-height: 150%;\"><i><span lang=\"et\"><span style=\"line-height: 150%;\">m <\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 150%;\">\u2013 aine mass\u00a0(\u00fchik gramm) <\/span><\/span><\/span><br><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">M<i> <\/i>\u2013 molaarmass (\u00fchik g\/mol)<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm; text-align: center;\">$n=\\frac{N}{N_{A}}$<\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span style=\"line-height: 150%;\"><i><span lang=\"et\"><span style=\"line-height: 150%;\">N\u00a0<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 150%;\">\u2013 osakeste arv (\u00fchik osakene)<\/span><\/span><\/span><br><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">N<sub>A<\/sub> \u2013 Avogadro arv, mille v\u00e4\u00e4rtus on 6,02 <\/span><\/span><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 150%;\">\u00d7 <\/span><\/span><\/span><span lang=\"et\"><span style=\"line-height: 150%;\">10<sup>23<\/sup> osakest\/mol<\/span><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm; text-align: center;\">$n=\\frac{V}{V_{m}}$<\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span style=\"line-height: 150%;\"><i><span lang=\"et\"><span style=\"line-height: 150%;\">V\u00a0<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 150%;\">\u2013 aine ruumala (\u00fchik dm<sup>3<\/sup>)<\/span><\/span><\/span><br><span style=\"line-height: 150%;\"><i><span lang=\"et\"><span style=\"line-height: 150%;\">V<\/span><\/span><\/i><sub><span lang=\"et\"><span style=\"line-height: 150%;\">m<\/span><\/span><\/sub><span lang=\"et\"><span style=\"line-height: 150%;\"> \u2013 molaarruumala (v\u00e4\u00e4rtus 22,4 dm<sup>3<\/sup>\/mol normaaltingimustel)<\/span><\/span><\/span><\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Tiheduse kaudu lahuse ruumala arvutamine:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm; text-align: center;\">$V=\\frac{m}{\\rho}$<\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span style=\"line-height: 150%;\"><i><span lang=\"et\"><span style=\"line-height: 150%;\">m <\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 150%;\">\u2013 kogu lahuse mass, <i>m<\/i>(lahus) = <i>m<\/i>(aine) + <i>m<\/i>(lahusti) (\u00fchik gramm)<\/span><\/span><\/span><br><i><span lang=\"et\"><span style=\"line-height: 115%;\">\u03c1 <\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">\u2013 lahuse tihedus (\u00fchik g\/cm<sup>3<\/sup>)<\/span><\/span>.<\/p>\n<h4>N\u00e4idis\u00fclesanne 1.\u00a0Molaarse kontsentratsiooni arvutamine<\/h4>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\"><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">250 cm<sup>3<\/sup> lahust sisaldab 10 g NaOH-d. Arvuta lahuse molaarne kontsentratsioon.<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12pt 0cm;\"><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">Arvuta <u>250 <\/u>cm<sup>3<\/sup> lahust sisaldab <u>10<\/u> g NaOH-d. Arvuta lahuse molaarne kontsentratsioon.<\/span><\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/td>\n<td><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">250 cm<sup>3<\/sup> on lahuse ruumala, 10 g on aine mass.<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(lahus) = 250 cm<sup>3<\/sup> <\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">m<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(aine) = 10 g<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">c\u00a0<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">= ?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">5. Kirjuta v\u00e4lja valemid ning vajadusel avalda valemist otsitav suurus.<\/span><\/span><\/p>\n<\/td>\n<td>\n<p>$c=\\frac{n}{V}$<\/p>\n<p><span lang=\"et\"><span style=\"line-height: 115%;\">Ainehulga saad arvutada valemist\u00a0<\/span><\/span>$n=\\frac{m}{M}$<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">6. Asenda valemitesse arvud ning arvuta.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Ainehulga arvutamiseks leia\u00a0k\u00f5igepealt NaOH molaarmass:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">M<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(NaOH) = 23 + 1 + 16 = 40 g\/mol<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$n=\\frac{m}{M}=\\frac{10g}{40g\/mol}=0,25 mol$<\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Molaarse kontsentratsiooni valemis peab lahuse ruumala olema dm<sup>3<\/sup>:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(lahus) = 250 cm<sup>3<\/sup> = 0,250 dm<sup>3<\/sup><\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\">$c=\\frac{n}{V}=\\frac{0,25 mol}{0,25 dm^{3}}=1\\frac{mol}{dm^{3}}=1 M$<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"margin: 12pt 0cm;\"><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801b9-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801b9-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801b9-collapse\">\u00dclesanne 1<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801b9-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801b9-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p style=\"margin: 12pt 0cm;\"><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801d3-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801d3-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801d3-collapse\">\u00dclesande 1 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801d3-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801d3-heading\">\n            <div class=\"accordion-body\">\n<p style=\"list-style-type: none;\">1.1 Arvuta lahuse molaarne kontsentratsioon, kui 1,50 l lahusesse on lisatud 1,45 mol KCl. Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus:<\/b> 0,97 M<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Lahendus:<\/b> Teada on ruumala <i>V<\/i> = 1,50 l ja moolide arv <i>n<\/i> = 1,45 mol. Neist saame arvutada molaarse kontsentratsiooni <i>c<\/i>:<\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\"><span class=\"math-tex\">$c = {n \\over V}={1.45 mol \\over 1.50l}\u22480,97 M$<\/span><\/p>\n<p style=\"list-style-type: none; margin-left: 80px;\">\n<\/p><p style=\"list-style-type: none;\">1.2 Arvuta lahuse molaarne kontsentratsioon, kui 10 ml lahusesse on lisatud 5,65 mmol Br<sub>2<\/sub>. Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus: <\/b>0,57 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on ruumala <i>V<\/i> = 10 ml ja moolide arv <i>n<\/i> = 5,65 mmol. Neist saame arvutada molaarse kontsentratsiooni <i>c<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c = {n \\over V}={5.65mmol \\over 10ml}={5.65mol \\over 10l}\u22480,57 M$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>1.3 Arvuta lahuse molaarne kontsentratsioon, kui 1575 l lahusesse on lisatud 20,54 kg Al(NO<sub>3<\/sub>)<sub>3<\/sub>. Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"list-style-type: none; margin-left: 40px;\"><b>Vastus:<\/b> 0,06 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on ruumala <i>V<\/i> = 1575 l ja aine mass <i>m<\/i> = 20,54 kg. Molaarse kontsentratsiooni leidmiseks tuleb esmalt arvutada moolide arv <i>n<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><i>M<\/i>[Al(NO<sub>3<\/sub>)<sub>3<\/sub>] = 213 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={20540g \\over 213g\/mol}=96.43mol$<\/span><i><\/i><\/p>\n<p style=\"margin-left: 40px;\">N\u00fc\u00fcd saab leida molaarse kontsentratsiooni <i>c<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c = {n \\over V}={96.43mol \\over 1575l}\u22480,06 M$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>1.4 Arvuta lahuse molaarne kontsentratsioon, kui 1 l lahust sisaldas 0,515 g v\u00e4\u00e4velhapet. Anna vastus t\u00e4psusega kolm kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,005 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on ruumala <i>V<\/i> = 1 l ja aine mass <i>m<\/i> = 0,515 g. Molaarse kontsentratsiooni leidmiseks tuleb esmalt arvutada moolide arv <i>n<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><i>M<\/i>(H<sub>2<\/sub>SO<sub>4<\/sub>) = 98 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={0.515g \\over 98g\/mol}=0.005mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">N\u00fc\u00fcd saab leida molaarse kontsentratsiooni <i>c<\/i>:<\/p>\n<p style=\"margin-left: 54.0pt;\"><span class=\"math-tex\">$c = {n \\over V}={0.005mol \\over 1l}=0,005 M$<\/span><\/p>\n<p style=\"margin-left: 54.0pt;\">\n<\/p><p>1.5 1 l piimas on 1,0 g Ca<sup>2+<\/sup> ioone. Mis on Ca<sup>2+<\/sup> molaarne kontsentratsioon piimas? Anna vastus t\u00e4psusega kolm kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,025 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on ruumala <i>V<\/i> = 1 l ja Ca<sup>2+<\/sup> ioonide mass <i>m<\/i> = 1,0 g. Molaarse kontsentratsiooni leidmiseks tuleb esmalt arvutada moolide arv <i>n:<\/i><\/p>\n<p style=\"margin-left: 80px;\"><i>M<\/i>(Ca<sup>2+<\/sup>) = 40 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={1.0g \\over 40g\/mol}=0.025mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">N\u00fc\u00fcd saab leida molaarse kontsentratsiooni <i>c<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c = {n \\over V}={0.025mol \\over 1l}=0,025 M$<\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h4>N\u00e4idis\u00fclesanne 2. Lahuse ruumala arvutamine<\/h4>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/td>\n<td>Mitu ml 3,5 M soolhappe lahust \u00a0tuleb v\u00f5tta, et see sisaldaks \u00a05 g puhast HCl-i?<\/td>\n<\/tr>\n<tr>\n<td>2. Jooni tekstis alla k\u00f5ik arvud.<\/td>\n<td>Mitu ml <u>3,5 M<\/u> soolhappe lahust \u00a0tuleb v\u00f5tta, et see sisaldaks \u00a0<u>5 g<\/u> puhast HCl-i?<\/td>\n<\/tr>\n<tr>\n<td>3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/td>\n<td>3,5 M on lahuse molaarsus, 5 g on puhta HCl mass.<\/td>\n<\/tr>\n<tr>\n<td>4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/td>\n<td>c = 3,5 M<br>m(aine) = 5 g<br>V(lahus) = ?<\/td>\n<\/tr>\n<tr>\n<td>5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda otsitav suurus.<\/td>\n<td>\n<p>$c=\\frac{n}{V}$<\/p>\n<p>Avalda\u00a0valemist lahuse ruumala\u00a0 $V=\\frac{n}{c}$.<br>Lahuse molaarne kontsentratsioon on teada, aga puudub ainehulk, mille saab arvutada\u00a0valemist $n=\\frac{m}{M}$. Selleks tuleb eelnevalt arvutada aine molaarmass.<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>6. Asenda valemitesse arvud ning arvuta.<\/td>\n<td>\n<p>M(HCl) = 1 + 35,5 = 36,5 g\/mol<\/p>\n<p>$n=\\frac{m}{M}=\\frac{5 g}{36,5 g\/mol}=0,14 mol$<\/p>\n<p>$V=\\frac{n}{c}=\\frac{0,14 mol}{3,5 mol\/dm^{3}}=0,04 dm^{3}=40 cm^{3}(ml)$<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801df-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801df-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801df-collapse\">\u00dclesanne 2<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801df-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801df-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801e5-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801e5-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801e5-collapse\">\u00dclesande 2 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801e5-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801e5-heading\">\n            <div class=\"accordion-body\">\n<p>2.1. 1,0 M Fe(NO<sub>3<\/sub>)<sub>3<\/sub> lahus sisaldab 10 g ainet. Arvuta lahuse ruumala milliliitrites. Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>41 ml<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on aine mass <i>m<\/i> = 10 g ja selle molaarne kontsentratsioon <i>c<\/i> = 1,0 M. Kui kontsentratsioon on teada, siis ruumala leidmiseks on veel vaja teada moolide arvu. Seda saab Fe(NO<sub>3<\/sub>)<sub>3<\/sub> massi abil leida:<\/p>\n<p style=\"margin-left: 80px;\"><i>M<\/i>[Fe(NO<sub>3<\/sub>)<sub>3<\/sub>] = 242 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={10g \\over 242g\/mol}=0.041mol$<\/span><\/p>\n<p style=\"margin-left: 40px;\">N\u00fc\u00fcd saab leida lahuse ruumala <i>V<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$V = {n \\over c}={0.041mol \\over 1.0M}=0.041l =41ml$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>2.2. 1,22 M lahuse lahjendamisel saadi 0,17 liitrit 0,36 M C<sub>3<\/sub>H<sub>7<\/sub>OH lahust. Arvuta esialgse lahuse ruumala? Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,050 l<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus: <\/b>Lahuste lahjendamise \u00fclesannetes on abiks asjaolu, et lahjendamisel lisatakse lahustit, kuid aine (selles \u00fclesandes C<sub>3<\/sub>H<sub>7<\/sub>OH) moolide arv <i>n<\/i> j\u00e4\u00e4b m\u00f5lemas lahuses samaks. Seega tuleb leida moolide arv l\u00f5pplahuses ja hiljem selle alusel arvutada alglahuse ruumala.<\/p>\n<p style=\"margin-left: 40px;\">L\u00f5pplahuse andmed on teada: <i>V<sub>l\u00f5pp<\/sub> <\/i>= 0,17 l, <i>c<sub>l\u00f5pp<\/sub><\/i> = 0,36 M; leiame moolide arvu <i>n<\/i>:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n\u00a0=\u00a0<\/span><\/i><i><span lang=\"ET\">V<\/span><\/i><i><span lang=\"ET\"><sub>l<\/sub><sub>\u00f5<\/sub><sub>pp<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">c<\/span><\/i><i><span lang=\"ET\"><sub>l<\/sub><sub>\u00f5<\/sub><sub>pp<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">0,17 l \u00d7 0,36 M = 0,0612 mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kuna see on moolide arv ka alglahuses, saame n\u00fc\u00fcd arvutada esialgse lahuse ruumala:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$V_{alg} = {n \\over c_{alg}}={0.0612mol \\over 1.22M}=0.05l $<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>2.3. 15 mg Na<sub>2<\/sub>SO<sub>4<\/sub> sisaldus 0,6 M lahuses. Kui suur on lahuse ruumala milliliitrites? Anna vastus t\u00e4psusega kolm kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,176 ml<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus<\/b>: Teada on kontsentratsioon <i>c<\/i> = 0,6 M ja aine mass <i>m<\/i> = 15 mg. Lahuse ruumala leidmiseks tuleb esmalt arvutada moolide arv <i>n<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><i>M<\/i>(Na<sub>2<\/sub>SO<sub>4<\/sub>) = 142 g\/mol = 142 mg\/mmol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={15mg \\over 142g\/mol}=0.1056mmol$<\/span><\/p>\n<p style=\"margin-left: 36.0pt;\">N\u00fc\u00fcd saab leida lahuse ruumala <i>V<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$V = {n \\over c}={0.1056mmol \\over 0.6mmol\/ml}=0.176ml$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>2.4. Mitmes ml lahuses peaks lahustama 250 kg NaNO<sub>3<\/sub>, et lahuse kontsentratsioon oleks 5 M? Anna vastus t\u00e4isarvuna.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>588235 ml<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on NaNO<sub>3<\/sub> mass <i>m<\/i> = 250 kg ja soovitav kontsentratsioon <i>c<\/i> = 5 M. Lahuse ruumala leidmiseks tuleb j\u00e4llegi esmalt arvutada moolide arv <i>n<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><i>M<\/i>(NaNO<sub>3<\/sub>) = 85 g\/mol<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$n = {m \\over M}={250000g \\over 85g\/mol}=2941.1765mol$<\/span><\/p>\n<p style=\"margin-left: 36.0pt;\">N\u00fc\u00fcd saab leida lahuse ruumala <i>V<\/i>:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$V = {n \\over c}={2941.1765mol \\over 5M}=588.235l=588235ml$<\/span><\/p>\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h4>N\u00e4idis\u00fclesanne 3. Ainehulga arvutamine<\/h4>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\"><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">Mitu mooli v\u00e4\u00e4velhapet sisaldub 500 ml 0,15 M H<sub>2<\/sub>SO<sub>4<\/sub> lahuses?<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/td>\n<td><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">Mitu mooli v\u00e4\u00e4velhapet sisaldub <u>500 <\/u>ml <u>0,15 <\/u>M H<sub>2<\/sub>SO<sub>4<\/sub> lahuses?<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/td>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">500 ml on kogu lahuse ruumala, 0,15 M on lahuse molaarne kontsentratsioon.<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(lahus) = 500 ml<\/span><\/span><\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">c\u00a0<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">= 0,15 M<\/span><\/span><\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">n<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(H<sub>2<\/sub>SO<sub>4<\/sub>) = ?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/span><\/span><\/td>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">Lahuse molaarsest kontsentratsioonist saame avaldada otsitava suuruse, ainehulga <i>n <\/i>= <i>c <\/i><\/span><\/span><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">\u00d7 <\/span><\/span><\/span><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">.<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">6. Asenda valemitesse arvud ning arvuta.<\/span><\/span><\/td>\n<td>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Teisenda lahuse ruumala dm<sup>3<\/sup>-teks <i>V<\/i>(lahus) = 500 ml = 0,5 <span style=\"background: white;\">dm<sup>3<\/sup><\/span><\/span><\/span><\/p>\n<p><i><span lang=\"et\"><span style=\"line-height: 115%;\">n <\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">= <i>c <\/i><\/span><\/span><span lang=\"et\"><span style=\"background: white;\"><span style=\"line-height: 115%;\">\u00d7 <\/span><\/span><\/span><i><span lang=\"et\"><span style=\"line-height: 115%;\">V <\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">= 0,15 mol\/dm<sup>3<\/sup> <span style=\"background: white;\">\u00d7 0,5 dm<sup>3\u00a0<\/sup>= 0,075 mol<\/span><\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"margin: 12pt 0cm;\"><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801eb-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801eb-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801eb-collapse\">\u00dclesanne 3<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801eb-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801eb-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n\n<p style=\"margin: 12pt 0cm\">\n\t<\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801f5-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801f5-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801f5-collapse\">\u00dclesande 3 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801f5-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801f5-heading\">\n            <div class=\"accordion-body\">\n\n<p style=\"margin: 12pt 0cm\">\n\t3.1. Mitu mooli ja mitu grammi ammooniumsulfaati sisaldub 150 cm<sup>3<\/sup> 0.5 M (NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub> lahuses? Vastust \u00e4ra \u00fcmarda.\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\t<b>Vastus: <\/b>0,075 mol ja 9,9 g\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\t<b>Lahendus<\/b>: Teada on ruumala <i>V<\/i> = 150 cm<sup>3<\/sup> ja kontsentratsioon <i>c<\/i> = 0,5 M. Neist saame arvutada moolide arvu <i>n<\/i>:<i><span lang=\"ET\" style=\",serif\"><\/span><\/i>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t<i><span lang=\"ET\" style=\",serif\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span style=\",serif\"><\/span><span lang=\"ET\" style=\",serif\">0,15 dm<\/span><span lang=\"ET\" style=\",serif\"><sup>3<\/sup>\u00a0<\/span><span lang=\"ET\" style=\",serif\">\u00d7 0,5 M = 0,075 mol<\/span>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\tSeda teades saab arvutada (NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub> massi:\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\tM[(NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub>] = 132 g\/mol<i><span lang=\"ET\" style=\",serif\"><\/span><\/i>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t<i><span lang=\"ET\" style=\",serif\">m\u00a0=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\" style=\",serif\">0,075 mol \u00d7 132 g\/mol = 9,9 g<\/span>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t\u00a0\n<\/p>\n<p style=\"margin-top: 12pt;margin-right: 0cm;margin-bottom: 12pt\">\n\t3.2. Kui suur on ainehulk, mis on lahustatud 25 ml 0,4 mol\/dm<sup>3<\/sup> lahuses?\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\t<b>Vastus:<\/b> 0,01 mol\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\t<b>Lahendus:<\/b> Teada on ruumala <i>V<\/i> = 25 ml ja molaarne kontsentratsioon <i>c<\/i> = 0,4 mol\/dm<sup>3<\/sup>. Neist saame arvutada ainehulga <i>n<\/i>:<i><span lang=\"ET\" style=\",serif\"><\/span><\/i>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t<i><span lang=\"ET\" style=\",serif\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span style=\",serif\"><\/span><span lang=\"ET\" style=\",serif\">25 cm<\/span><span lang=\"ET\" style=\",serif\"><sup>3<\/sup>\u00a0<\/span><span lang=\"ET\" style=\",serif\">\u00d7 0,4 mol\/<\/span><span style=\",serif\"><\/span><span lang=\"ET\" style=\",serif\">dm<\/span><span lang=\"ET\" style=\",serif\"><sup>3<\/sup>\u00a0<\/span><span lang=\"ET\" style=\",serif\">=\u00a0<\/span><span style=\",serif\"><\/span><span lang=\"ET\" style=\",serif\">0,025 dm<\/span><span lang=\"ET\" style=\",serif\"><sup>3<\/sup>\u00a0<\/span><span lang=\"ET\" style=\",serif\">\u00d7 0,4 mol\/<\/span><span style=\",serif\"><\/span><span lang=\"ET\" style=\",serif\">dm<\/span><span lang=\"ET\" style=\",serif\"><sup>3<\/sup>\u00a0<\/span><span lang=\"ET\" style=\",serif\">= 0,01 mol<\/span>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t\u00a0\n<\/p>\n<p style=\"margin-top: 12pt;margin-right: 0cm;margin-bottom: 12pt\">\n\t3.3. Mitu mooli v\u00e4\u00e4velhapet on 2 l 18,5 M v\u00e4\u00e4velhappe lahuses?\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\t<b>Vastus: <\/b>37 mol\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 40px\">\n\t<b>Lahendus:<\/b>Teada on ruumala <i>V<\/i> = 2 l ja molaarne kontsentratsioon <i>c<\/i> = 18,5 M. Neist saame arvutada moolide arvu <i>n<\/i>:<i><span lang=\"ET\" style=\",serif\"><\/span><\/i>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t<i><span lang=\"ET\" style=\",serif\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\" style=\",serif\">2 l \u00d7 18,5 mol\/l = 37 mol<\/span>\n<\/p>\n<p style=\"margin: 12pt 0cm 12pt 80px\">\n\t\u00a0\n<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t3.4. Mitu mooli FeSO<sub>4<\/sub> on 325 ml 1,8 \u00d7 10<sup>\u22126<\/sup> M FeSO<sub>4<\/sub> lahuses? Vastust \u00e4ra \u00fcmarda.\n<\/p>\n<p style=\"margin-left: 40px\">\n\t<b>Vastus:<\/b> 0,000000585 mol\n<\/p>\n<p style=\"margin-left: 40px\">\n\t<b>Lahendus: <\/b>Teada on lahuse ruumala <i>V<\/i> = 325 ml ja molaarne kontsentratsioon <i>c<\/i> = 1,8\u00d710<sup>\u22126<\/sup> M. Neist saame arvutada moolide arvu <i>n<\/i>:<i><span lang=\"ET\" style=\",serif\"><\/span><\/i>\n<\/p>\n<p style=\"margin-left: 80px\">\n\t<i><span lang=\"ET\" style=\",serif\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\" style=\",serif\">0,325 l \u00d7\u00a0<\/span><span style=\",serif\">1,8 \u00d7\u00a0<\/span><span style=\",serif\"><\/span><span style=\",serif\">10<\/span><sup><span style=\",serif\"><sup>\u2013<\/sup><sup>6<\/sup><\/span><\/sup><span lang=\"ET\" style=\",serif\"> mol\/l = 0,000000585 mol<\/span>\n<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t<\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n\n<h4>\n\tN\u00e4idis\u00fclesanne 4. Ioonide kontsentratsioonide arvutamine<br>\n<\/h4>\n<table class=\"table table-hover\" border=\"0\" cellpadding=\"1\" cellspacing=\"1\" style=\"width: 100%\">\n<tbody>\n<tr>\n<td style=\"width: 30%\">\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"background:white\"><span style=\"line-height:115%\"><span style=\",serif\">CaCl<sub>2<\/sub> on tugev elektrol\u00fc\u00fct ning dissotsieerub t\u00e4ielikult. 50 ml CaCl<sub>2<\/sub> lahuses oli lahustatud 0,2 mol CaCl<sub>2<\/sub>. Arvuta lahusesse tekkinud Ca<sup>2+<\/sup> ja Cl<sup>\u2013<\/sup> ioonide kontsentratsioon lahuses.<\/span><\/span><\/span><\/span>\n\t\t\t<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"background:white\"><span style=\"line-height:115%\"><span style=\",serif\">CaCl<sub>2<\/sub> on tugev elektrol\u00fc\u00fct ning dissotsieerub\u00a0<\/span><\/span><\/span><\/span><span lang=\"et\"><span style=\"background:white\"><span style=\"line-height:115%\"><span style=\",serif\">t\u00e4ielikult. <u>50<\/u> ml CaCl<sub>2 <\/sub>lahuses oli lahustatud <u>0,2<\/u> mol CaCl<sub>2<\/sub>. Aruta lahusesse tekkinud Ca<sup>2+<\/sup> ja Cl<sup>\u2013<\/sup> ioonide kontsentratsioon lahuses.<\/span><\/span><\/span><\/span>\n\t\t\t<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">50 ml on kogu lahuse ruumala, 0,2 mol on lahuses sisalduv <span style=\"background:white\">CaCl<sub>2<\/sub> aine hulk.<\/span><\/span><\/span><\/span>\n\t\t\t<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">V<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(lahus) = 50 ml<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">n <\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">= 0,2 mol<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">c<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(Ca<sup>2+<\/sup>) =?\u00a0<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">c<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(Cl<sup>\u2013<\/sup>) = ?<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n\t\t\t\t$c=\\frac{n}{V}$\n\t\t\t<\/td>\n<\/tr>\n<tr>\n<td>\n\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">6. Asenda valemitesse arvud ning arvuta.<\/span><\/span><\/span>\n\t\t\t<\/td>\n<td>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Kirjutame v\u00e4lja dissotsiatsiooniv\u00f5rrandi:<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"background:white\"><span style=\"line-height:115%\"><span style=\",serif\">CaCl<sub>2<\/sub><\/span><\/span><\/span><\/span><span lang=\"et\"><span style=\"background:white\"><span style=\"line-height:115%\"><span> \u2192 <\/span><\/span><\/span><\/span><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Ca<sup>2+<\/sup> + 2 Cl<sup>\u2013<\/sup><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">V\u00f5rrandist on n\u00e4ha, et \u00fche mooli <span style=\"background:white\">CaCl<sub>2<\/sub> dissotsieerumisel tekib 1 mool\u00a0 <\/span>Ca<sup>2+<\/sup> ja 2 mooli Cl<sup>\u2013<\/sup>. \u00dclesande alusel lisati lahusesse 0,2 mooli soola, seega saab tekkida 1 <span style=\"background:white\">$\\times $ <\/span>0,2 mooli Ca<sup>2+<\/sup> ja 2\u00a0$\\times $\u00a0<span style=\"background:white\">0,2 <\/span>\u00a0mooli Cl<sup>\u2013<\/sup>. <\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">Enne molaarse kontsentratsiooni arvutamist teisendame lahuse ruumala.<\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t<i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">V<\/span><\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height:115%\"><span style=\",serif\">(lahus) = 50 ml = 0,05 <span style=\"background:white\">dm<sup>3\u00a0<\/sup><\/span><\/span><\/span><\/span>\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t$c(Ca^{2+})=\\frac{n}{V}=\\frac{0,2 mol}{0,05 dm^{3}}=4\\frac{mol}{dm^{3}}=4M$\n\t\t\t\t<\/p>\n<p style=\"margin: 12pt 0cm\">\n\t\t\t\t\t$c(Cl^{-})=\\frac{n}{V}=\\frac{0,4 mol}{0,05 dm^{3}}=8\\frac{mol}{dm^{3}}=8M$\n\t\t\t\t<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n\t<\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c801fd-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c801fd-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c801fd-collapse\">\u00dclesanne 4<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c801fd-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c801fd-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c80208-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c80208-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c80208-collapse\">\u00dclesande 4 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c80208-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c80208-heading\">\n            <div class=\"accordion-body\">\n<p>4.1. 500 ml lahusesse oli lisatud 0,2 mol CaCl<sub>2<\/sub> ja 0,7 mol HCl. Arvuta kloriidioonide ja vesinikioonide kontsentratsioon lahuses. Anna vastus t\u00e4psusega \u00fcks koht p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>2,2 M; 1,4 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on lahuse ruumala <i>V<\/i> = 500 ml, CaCl<sub>2<\/sub> moolide arv <i>n<\/i>(CaCl<sub>2<\/sub>) = 0,2 mol ja HCl moolide arv <i>n<\/i>(HCl) = 0,7 mol. Vaja on leida <i>c<\/i>(Cl<sup>\u2013<\/sup>) ja <i>c<\/i>(H<sup>+<\/sup>).<\/p>\n<p style=\"margin-left: 40px;\">Kirjutame v\u00e4lja kummagi \u00fchendi dissotsiatsiooniv\u00f5rrandid:<\/p>\n<p style=\"margin-left: 80px;\">1) CaCl<sub>2<\/sub> \u2192 Ca<sup>2+<\/sup> + 2Cl<sup>\u2013<\/sup><\/p>\n<p style=\"margin-left: 80px;\">2) HCl \u2192 H<sup>+<\/sup> + Cl<sup>\u2013<\/sup><\/p>\n<p style=\"margin-left: 40px;\">V\u00f5rrandist 1) n\u00e4eme, et 1 mooli CaCl<sub>2<\/sub> dissotsieerumisel tekib 2 mooli kloriidioone. Seega tekib 0,2 mol CaCl<sub>2<\/sub> dissotsiatsioonil 2\u00d70,2 = 0,4 mol kloriidioone.<\/p>\n<p style=\"margin-left: 40px;\">V\u00f5rrandi 2) p\u00f5hjal saame \u00f6elda, et HCl dissotsiatsioonil tekib 1\u00d70,7 = 0,7 mol kloriidioone. Vesinikioone tekib sama palju, seega <i>n<\/i>(H<sup>+<\/sup>) =<i> <\/i>0,7 mol.<\/p>\n<p style=\"margin-left: 40px;\">Kloriidioone on kokku <i>n<\/i>(Cl<sup>\u2013<\/sup>) = 0,4 + 0,7 = 1,1 mol.<\/p>\n<p style=\"margin-left: 40px;\">Ioonide ainehulki teades saame arvutada nende kontsentratsioonid lahuses.<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c(Cl^-)={n \\over V}={1.1mol \\over 0.5l}=2.2M$<\/span><\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c(H^+)={n \\over V}={0.7mol \\over 0.5l}=1.4M$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>4.2. Arvuta fosfaatioonide ja kaaliumioonide ainehulk ning molaarne kontsentratsioon 200 ml 0,4 M kaaliumfosfaadi t\u00e4ielikul dissotsiatsioonil. L\u00f5ppvastuseid \u00e4ra \u00fcmarda.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,24 mol; 1,2 M; 0,08 mol; 0,4 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus: <\/b>Teada on lahuse ruumala <i>V<\/i> = 200 ml ja K<sub>3<\/sub>PO<sub>4<\/sub> molaarne kontsentratsioon <i>c<\/i>(K<sub>3<\/sub>PO<sub>4<\/sub>) = 0,4 M. Neist saame leida K<sub>3<\/sub>PO<sub>4<\/sub> moolide arvu:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,2 l \u00d7 0,4 mol\/l = 0,08 mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kirjutame v\u00e4lja K<sub>3<\/sub>PO<sub>4<\/sub> dissotsiatsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">K<sub>3<\/sub>PO<sub>4<\/sub> \u2192 3K<sup>+<\/sup> + PO<sub>4<\/sub><sup>3-<\/sup><\/p>\n<p style=\"margin-left: 40px;\">V\u00f5rrandist n\u00e4eme, et dissotsiatsioonil tekkivate kaaliumioonide moolide arv on kolm korda suurem kui dissotsieerumata soola ja fosfaatioonide omad. Seega <i>n<\/i>(K<sup>+<\/sup>) = 3\u00d70,08 = 0,24 mol ja <i>n<\/i>(PO<sub>4<\/sub><sup>3-<\/sup>) = 0,08 mol.<\/p>\n<p style=\"margin-left: 40px;\">Ainehulki teades saame leida ka ioonide molaarsed kontsentratsioonid:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c(K^+)={n \\over V}={0.24mol \\over 0.2l}=1.2M$<\/span><\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c(PO^{-3}_4)={n \\over V}={0.08mol \\over 0.2l}=0.4M$<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>4.3. Arvuta kui palju bromiidioone ja magneesiumi ioone tekib 400 ml 0,15 M magneesiumbromiidi t\u00e4ielikul dissotsieerumisel. Anna vastus \u00fche koma koha t\u00e4psusega.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus:<\/b> N(Mg<sup>2+<\/sup>)= 3,6 \u00d7 10<sup>22<\/sup>,\u00a0N(Br<sup>\u2212<\/sup>)= 7,2 \u00d7 10<sup>22<\/sup><\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on lahuse ruumala <i>V<\/i> = 400 ml ja MgBr<sub>2<\/sub> molaarne kontsentratsioon <i>c<\/i>(MgBr<sub>2<\/sub>) = 0,15 M. Neist saame leida MgBr<sub>2<\/sub> moolide arvu:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,4 l \u00d7 0,15 mol\/l = 0,06 mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Kirjutame v\u00e4lja MgBr<sub>2<\/sub> dissotsiatsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">MgBr<sub>2<\/sub> \u2192 Mg<sup>2+<\/sup> + 2Br<sup>\u2013<\/sup><\/p>\n<p style=\"margin-left: 40px;\">V\u00f5rrandist n\u00e4eme, et dissotsiatsioonil tekkivate kloriidioonide moolide arv on kaks korda suurem kui dissotsieerumata soola ja magneesiumioonide omad. Seega <i>n<\/i>(Mg<sup>2+<\/sup>) = 1\u00d70,06 = 0,06 mol ja <i>n<\/i>(Br<sup>\u2013<\/sup>) = 2\u00d70,06 = 0,12 mol.<\/p>\n<p style=\"margin-left: 40px;\">Osakeste arvu leidmiseks tuleb ainehulgad l\u00e4bi korrutada Avogadro arvuga <i>N<\/i><sub>A<\/sub>:<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">N<\/span><\/i><span lang=\"ET\">(<\/span><span lang=\"ET\">Mg<\/span><sup><i><span lang=\"ET\"><sup>2+<\/sup><\/span><\/i><\/sup><span lang=\"ET\">)\u00a0<i>=\u00a0<\/i><i>n\u00a0<\/i><i>\u00d7\u00a0<\/i><\/span><i><span lang=\"ET\">N<\/span><\/i><i><span lang=\"ET\"><sub>A<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">= 0<\/span><\/i><span lang=\"ET\">,06 mol \u00d7\u00a0<\/span>6,02 \u00d7 10<sup>23<\/sup>\u00a0aatomit\/mol = 3,6\u00d710<sup><sup>22<\/sup><\/sup><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">N<\/span><\/i><span lang=\"ET\">(<\/span><span lang=\"ET\">Br<\/span><sup><i><span lang=\"ET\"><sup>\u2013<\/sup><\/span><\/i><\/sup><span lang=\"ET\">)\u00a0<i>=\u00a0<\/i><i>n\u00a0<\/i><i>\u00d7\u00a0<\/i><\/span><i><span lang=\"ET\">N<\/span><\/i><i><span lang=\"ET\"><sub>A<\/sub>\u00a0<\/span><\/i><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">0,12 mol \u00d7\u00a0<\/span>6,02 \u00d7 10<sup>23<\/sup>\u00a0aatomit\/mol = 7,2\u00d710<sup><sup>22<\/sup><\/sup><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>4.4. 0,250 l lahusesse oli lisatud 0,2 mol v\u00e4\u00e4velhapet ja 0,5 mol alumiiniumsulfaati. Arvuta alumiiniumi ioonide ja sulfaatioonide kontsentratsioon lahuses. L\u00f5ppvastuseid \u00e4ra \u00fcmarda.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>6,8 M; 4 M<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Teada on lahuse ruumala <i>V<\/i> = 0,25 l, H<sub>2<\/sub>SO<sub>4<\/sub> moolide arv <i>n<\/i>(H<sub>2<\/sub>SO<sub>4<\/sub>) = 0,2 mol ja Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub> moolide arv <i>n<\/i>[Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>] = 0,5 mol. Kirjutame v\u00e4lja kummagi \u00fchendi dissotsiatsiooniv\u00f5rrandid:<\/p>\n<p style=\"margin-left: 80px;\">1) H<sub>2<\/sub>SO<sub>4 <\/sub>\u2192 2H<sup>+<\/sup> + SO<sub>4<\/sub><sup>2\u2013<\/sup><\/p>\n<p style=\"margin-left: 80px;\">2) Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub> \u2192 2Al<sup>3+<\/sup> + 3SO<sub>4<\/sub><sup>2\u2013<\/sup><\/p>\n<p style=\"margin-left: 40px;\">V\u00f5rrandi 1) j\u00e4rgi n\u00e4eme, et kui <i>n<\/i>(H<sub>2<\/sub>SO<sub>4<\/sub>) = 0,2 mol, siis <i>n<\/i>(H<sup>+<\/sup>) = 2\u00d70,2 = 0,4 mol ja <i>n<\/i>(SO<sub>4<\/sub><sup>2\u2013<\/sup>) = 0,2 mol. V\u00f5rrandi 2) j\u00e4rgi n\u00e4eme, et kui <i>n<\/i>(Al<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3<\/sub>) = 0,5 mol, siis <i>n<\/i>(Al<sup>3+<\/sup>) = 2\u00d70,5 = 1,0 mol ja <i>n<\/i>(SO<sub>4<\/sub><sup>2\u2013<\/sup>) = 3\u00d70,5 = 1,5 mol.<\/p>\n<p style=\"margin-left: 40px;\">Sulfaatioone on kokku <i>n<\/i>(SO<sub>4<\/sub><sup>2\u2013<\/sup>) = 0,2 + 1,5 = 1,7 mol ja alumiiniumiioone <i>n<\/i>(Al<sup>3+<\/sup>) = 1,0 mol. N\u00fc\u00fcd saab leida kummagi molaarsed kontsentratsioonid:<\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c(Al^{3+})={n \\over V}={1.0mol \\over 0.25l}=4M$<\/span><\/p>\n<p style=\"margin-left: 80px;\"><span class=\"math-tex\">$c(SO^{2-}_4)={n \\over V}={1.7mol \\over 0.25l}=6.8M$<\/span><\/p>\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<h4>N\u00e4idis\u00fclesanne 5. Reaktsiooniv\u00f5rrandi abil ainehulga v\u00f5i ruumala arvutamine<\/h4>\n<table class=\"table table-hover\" style=\"width: 100%;\" border=\"0\" cellspacing=\"1\" cellpadding=\"1\">\n<tbody>\n<tr>\n<td style=\"width: 30%;\"><span lang=\"et\"><span style=\"line-height: 115%;\">1. Loe \u00fclesande tekst m\u00f5ttega l\u00e4bi.<\/span><\/span><\/td>\n<td><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">Mitu ml 1,5 M H<sub>2<\/sub>SO<sub>4<\/sub> lahust on vaja 2 M 15 ml NaOH lahuse neutraliseerimiseks?<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">2. Jooni tekstis alla k\u00f5ik arvud.<\/span><\/span><\/td>\n<td><span style=\"line-height: 150%;\"><span lang=\"et\"><span style=\"line-height: 150%;\">Mitu ml <u>1,5<\/u> M H<sub>2<\/sub>SO<sub>4<\/sub> lahust on vaja <u>2<\/u> M <u>15<\/u> ml NaOH lahuse neutraliseerimiseks?<\/span><\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">3. Omista arvule kindel f\u00fc\u00fcsikaline suurus. Seda on lihtne teha n\u00e4iteks \u00fchiku alusel.<\/span><\/span><\/td>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">1,5 M on v\u00e4\u00e4velhappe lahuse molaarne kontsentratsioon, 2 M on NaOH lahuse molaarne kontsentratsioon ja 15 ml on NaOH lahuse kogus.<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">4. Kirjuta v\u00e4lja andmed ning otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">c<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(H<sub>2<\/sub>SO<sub>4<\/sub>) = 1,5 M<\/span><\/span><\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">c<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(NaOH) = 2 M<\/span><\/span><\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(NaOH) = 15 ml = 0,015 dm<sup>3<\/sup><\/span><\/span><\/p>\n<p style=\"text-align: justify; margin: 12.0pt 0cm 12.0pt 0cm;\"><i><span lang=\"et\"><span style=\"line-height: 115%;\">V<\/span><\/span><\/i><span lang=\"et\"><span style=\"line-height: 115%;\">(H<sub>2<\/sub>SO<sub>4<\/sub>) = ?<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">5. Kirjuta v\u00e4lja valemid ning kui vaja, avalda valemist otsitav suurus.<\/span><\/span><\/td>\n<td>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Koostame reaktsiooniv\u00f5rrandi:<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">2NaOH + H<sub>2<\/sub>SO<sub>4<\/sub><\/span><\/span><span lang=\"et\"><span style=\"line-height: 115%;\"> \u2192 Na<\/span><\/span><sub><span lang=\"et\"><span style=\"line-height: 115%;\">2<\/span><\/span><\/sub><span lang=\"et\"><span style=\"line-height: 115%;\">SO<sub>4<\/sub> + 2H<sub>2<\/sub>O<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">Reaktsiooniv\u00f5rrandi alusel on n\u00e4ha, et 2 mooli NaOH-ga reageerib \u00fcks mool\u00a0 H<sub>2<\/sub>SO<sub>4<\/sub> .<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">N\u00fc\u00fcd vaatame, mitu mooli ainet on antud \u00fclesandes. Moolide arvu arvutame molaarse kontsentratsiooni ja aine ruumala alusel<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">$c=\\frac{n}{V}$ ja <i>n<\/i>(NaOH) = <i>c <\/i><\/span><span style=\"background: white;\"><span style=\"line-height: 115%;\">\u00d7 <\/span><\/span><\/span><i><span lang=\"et\"><span style=\"line-height: 115%;\">V <\/span><\/span><\/i><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">V\u00e4\u00e4velhappe ruumala arvutamiseks kasutame valemit<\/span><\/span><\/p>\n<p style=\"margin: 12pt 0cm;\"><span lang=\"et\"><span style=\"line-height: 115%;\">$V=\\frac{n}{c}$<\/span><\/span><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td><span lang=\"et\"><span style=\"line-height: 115%;\">6. Asenda valemitesse arvud ning arvuta.<\/span><\/span><\/td>\n<td>\n<p>Kirjutame reaktsiooniv\u00f5rrandi<\/p>\n<p>H<sub>2<\/sub>SO<sub>4<\/sub> + 2NaOH \u2192 Na<sub>2<\/sub>SO<sub>4<\/sub> + 2H<sub>2<\/sub>O<\/p>\n<p>H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0 ja NaOH moolsuhe on 1 : 2.<\/p>\n<p>\u00dclesandes antud andmete alusel<\/p>\n<p>n(NaOH) = c $\\times $ V = 2 mol\/dm<sup>3<\/sup> $\\times $ 0,015 dm<sup>3<\/sup> = 0,03 mol<\/p>\n<p>J\u00e4relikult v\u00e4\u00e4velhappe moolide arv on:<\/p>\n<p>$n(H_{2}SO_{4})=\\frac{1}{2}\\times 0,03 mol=0,015 mol$<\/p>\n<p>Arvutame v\u00e4\u00e4velhappe ruumala<\/p>\n<p>$V(H_{2}SO_{4})=\\frac{n}{c}=\\frac{0,015 mol}{1,5 mol\/dm^{3}}=0,01dm^{3}=10cm^{3}(ml)$<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"margin: 12pt 0cm;\"><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c8020d-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c8020d-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c8020d-collapse\">\u00dclesanne 5<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c8020d-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c8020d-heading\">\n            <div class=\"accordion-body\">\n\n\n<div class=\"h5p-iframe-wrapper\"><div class=\"video-placeholder-wrapper video-placeholder-wrapper--fixed\" style=\"height: 366px;\">\n\t\t\t    <div class=\"video-placeholder d-flex justify-content-center align-items-center\">\n\t\t\t        <div class=\"overlay text-white p-2 w-100 text-center d-block justify-content-center align-items-center\">\n\t\t\t            <div>Kolmandate osapoolte sisu n\u00e4gemiseks palun n\u00f5ustu k\u00fcpsistega.<\/div>\n\t\t\t            <button class=\"btn btn-secondary btn-sm mt-1 consent-change\">Muuda n\u00f5usolekut<\/button>\n\t\t\t        <\/div>\n\t\t\t    <\/div>\n\t\t\t<\/div>\n<\/div>\n\n\n<p><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>\n<p style=\"margin: 12pt 0cm;\"><\/p><div class=\"accordion mb-3\">\n        <div class=\"accordion-item accordion-item--white\">\n        <h2 class=\"accordion-header\" id=\"accordion-69dc324c80216-heading\">\n            <button class=\"accordion-button collapsed\" type=\"button\" data-bs-toggle=\"collapse\" data-bs-target=\"#accordion-69dc324c80216-collapse\" aria-expanded=\"true\" aria-controls=\"accordion-69dc324c80216-collapse\">\u00dclesande 5 lahendused<\/button>\n        <\/h2>\n        <div id=\"accordion-69dc324c80216-collapse\" class=\"accordion-collapse collapse\" aria-labelledby=\"accordion-69dc324c80216-heading\">\n            <div class=\"accordion-body\">\n<p>5.1. Mitu mooli ja mitu grammi kaaliumnitraati tekib, kui tahkele kaaliumkarbonaadile lisati 30 ml 0,2 M l\u00e4mmastikhappe lahust?<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,006 mol; 0,6 g<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame v\u00e4lja ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">K<sub>2<\/sub>CO<sub>3<\/sub> + 2 HNO<sub>3<\/sub> \u2192 2 KNO<sub>3<\/sub> + CO<sub>2<\/sub> + H<sub>2<\/sub>O<\/p>\n<p style=\"margin-left: 40px;\">Kuna teada on l\u00e4mmastikhappe lahuse ruumala <i>V<\/i> = 30 ml ja selle kontsentratsioon <i>c<\/i>(HNO<sub>3<\/sub>) = 0,2 M, saame leida selle moolide arvu <i>n<\/i>:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,03 l \u00d7 0,2 mol\/l = 0,006 mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandist n\u00e4eme, et HNO<sub>3<\/sub> ja KNO<sub>3<\/sub> moolsuhe on 1:1. Seega tekib ka KNO<sub>3<\/sub> 0,006 mol. N\u00fc\u00fcd saame arvutada molaarmassi ja massi:<\/p>\n<p style=\"margin-left: 80px;\">M(KNO<sub>3<\/sub>) = 101 g\/mol<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m\u00a0=\u00a0n\u00a0\u00d7\u00a0M=\u00a0<\/span><\/i><span lang=\"ET\">0,006 mol \u00d7 101 g\/mol = 0,606 g \u2248 0,6 g<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>5.2. Valati kokku 50 cm<sup>3 <\/sup>1,5 M kaaliumh\u00fcdroksiid\u00a0lahust ja 0,035 dm<sup>3<\/sup> 2 M fosforhappe lahust. Mitu mooli ja grammi kaali<i><span lang=\"ET\" style=\"color: red;\">u<\/span><\/i>mfosfaati tekib?<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus:<\/b> 0,025 mol; 5,3 g.<\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus: <\/b>Kirjutame v\u00e4lja ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">3 KOH + H<sub>3<\/sub>PO<sub>4<\/sub> \u2192 K<sub>3<\/sub>PO<sub>4<\/sub> + 3 H<sub>2<\/sub>O<\/p>\n<p style=\"margin-left: 40px;\">Olemasolevate andmete p\u00f5hjal tuleb leida reageerivate kaaliumh\u00fcdroksiidi ja fosforhappe ainehulgad.<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n<\/span><\/i><span lang=\"ET\">(KOH)\u00a0<i>=\u00a0<\/i><\/span><i><span lang=\"ET\">V<\/span><\/i><span lang=\"ET\"><sub>KOH<\/sub>\u00a0<\/span><i><span lang=\"ET\">\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">c<\/span><\/i><span lang=\"ET\"><sub>KOH<\/sub>\u00a0<\/span><i><span lang=\"ET\">= 0,05<\/span><\/i><span lang=\"ET\"> l\u00a0<i>\u00d7 1,5 <\/i>M\u00a0<i>= 0,075 <\/i>mol<\/span><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n<\/span><\/i><span lang=\"ET\">(<\/span><span lang=\"ET\">H<\/span><sub><span lang=\"ET\"><sub>3<\/sub><\/span><\/sub><span lang=\"ET\">PO<\/span><sub><span lang=\"ET\"><sub>4<\/sub><\/span><\/sub><span lang=\"ET\">)\u00a0<i>=\u00a0<\/i><\/span><i><span lang=\"ET\">V<\/span><\/i><sub><span lang=\"ET\"><sub>H<\/sub><\/span><\/sub><sub><span lang=\"ET\"><sub>3<\/sub><\/span><span lang=\"ET\"><sub>4<\/sub><\/span><span lang=\"ET\"><sub>PO<\/sub><\/span><span lang=\"ET\"><sub>\u00a0<\/sub><\/span><\/sub><i><span lang=\"ET\">\u00d7\u00a0<\/span><\/i><i><span lang=\"ET\">c<\/span><\/i><sub><span lang=\"ET\"><sub>H<\/sub><\/span><span lang=\"ET\"><sub>3<\/sub><\/span><span lang=\"ET\"><sub>4<\/sub><\/span><span lang=\"ET\"><sub>PO<\/sub><\/span><span lang=\"ET\"><sub>\u00a0<\/sub><\/span><\/sub><i><span lang=\"ET\">=\u00a0<\/span><\/i><span lang=\"ET\">0,035 l\u00a0<i>\u00d7 2 <\/i>M\u00a0<i>= 0,07 <\/i>mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandi j\u00e4rgi on KOH ja H<sub>3<\/sub>PO<sub>4<\/sub> moolsuhe 3:1, kokku valatud ainehulkade suhe aga 0,075:0,07. Seega on H<sub>3<\/sub>PO<sub>4<\/sub> \u00fclehulgas ja arvutused teeme KOH j\u00e4rgi.<\/p>\n<p style=\"margin-left: 40px;\">Kui 3 mooli KOH reageerimisel tekib 1 mool K<sub>3<\/sub>PO<sub>4<\/sub>, siis 0,075 mol KOH puhul tekib <i>n<\/i>=0,075\/3=0,025 mol K<sub>3<\/sub>PO<sub>4<\/sub>. Selle ainehulga massi saab juba leida:<\/p>\n<p style=\"margin-left: 80px;\">M(K<sub>3<\/sub>PO<sub>4<\/sub>) = 212 g\/mol<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">m\u00a0=\u00a0n\u00a0\u00d7\u00a0M\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,025 mol \u00d7 212 g\/mol = 5,3 g<\/span><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>5.3. Mitu dm<sup>3<\/sup> 1,5 M kaaliumh\u00fcdroksiidi lahust on vaja v\u00f5tta, et t\u00e4ielikult neutraliseerida 0,400 dm<sup>3<\/sup> 1 M l\u00e4mmastikhappe lahus? Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,27 dm<sup>3<\/sup><\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus<\/b>: Kirjutame v\u00e4lja ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">KOH + HNO<sub>3<\/sub> \u2192 KNO<sub>3<\/sub> + H<sub>2<\/sub>O<\/p>\n<p style=\"margin-left: 40px;\">Kuna teada on l\u00e4mmastikhappe lahuse ruumala <i>V<\/i> = 0,4 dm<sup>3<\/sup> ja selle kontsentratsioon <i>c<\/i>(HNO<sub>3<\/sub>) = 1 M, saame leida moolide arvu <i>n<\/i>:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,4 l \u00d7 1 mol\/l = 0,4 mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandi j\u00e4rgi reageerivad KOH ja HNO<sub>3<\/sub> moolsuhtega 1:1, seega on ka KOH moolide arv 0,4 mol. Kuna teame selle molaarset kontsentratsiooni <i>c<\/i>(KOH) = 1,5 M, saame arvutada KOH lahuse ruumala:<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\"><span class=\"math-tex\">$V = {n \\over c}={0.4mol \\over 1.5M}=0.266\u2026\u2248 0.27dm^3$<\/span><\/span><\/i><\/p>\n<p style=\"margin-left: 80px;\">\n<\/p><p>5.4. Mitu dm<sup>3<\/sup> 0,3 M baariumh\u00fcdroksiidi lahust on vaja v\u00f5tta, et t\u00e4ielikult neutraliseerida 0,25 dm<sup>3<\/sup> 0,25 M vesinikkloriidhappe lahus? Anna vastus t\u00e4psusega kaks kohta p\u00e4rast koma.<\/p>\n<p style=\"margin-left: 40px;\"><b>Vastus: <\/b>0,10 dm<sup>3<\/sup><\/p>\n<p style=\"margin-left: 40px;\"><b>Lahendus:<\/b> Kirjutame v\u00e4lja ja tasakaalustame reaktsiooniv\u00f5rrandi:<\/p>\n<p style=\"margin-left: 80px;\">Ba(OH)<sub>2<\/sub> + 2 HCl \u2192 BaCl<sub>2<\/sub> + 2 H<sub>2<\/sub>O<\/p>\n<p style=\"margin-left: 40px;\">Kuna teada on vesinikkloriidhappe lahuse ruumala <i>V<\/i> = 0,25 dm<sup>3<\/sup> ja selle kontsentratsioon <i>c<\/i>(HCl) = 0,25 M, saame leida moolide arvu <i>n<\/i>:<i><\/i><\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\">n\u00a0=\u00a0V\u00a0\u00d7\u00a0c\u00a0=\u00a0<\/span><\/i><span lang=\"ET\">0,25 l \u00d7 0,25 mol\/l = 0,0625 mol<\/span><\/p>\n<p style=\"margin-left: 40px;\">Reaktsiooniv\u00f5rrandi j\u00e4rgi reageerivad Ba(OH)<sub>2<\/sub> ja HCl moolsuhtega 1:2, seega on ka Ba(OH)<sub>2<\/sub> moolide arv 0,0625\/2=0,03125 mol. Kuna teame selle molaarset kontsentratsiooni <i>c<\/i>[Ba(OH)<sub>2<\/sub>] = 0,3 M, saame arvutada KOH lahuse ruumala:<\/p>\n<p style=\"margin-left: 80px;\"><i><span lang=\"ET\"><span class=\"math-tex\">$V = {n \\over c}={0.03125mol \\over 0.3M}\u22480.10dm^3$<\/span><\/span><\/i><\/p>\n<p style=\"margin: 12pt 0cm;\"><\/p><\/div>\n        <\/div>\n        <\/div>\n    <\/div>","protected":false},"excerpt":{"rendered":"<p>Nagu sa juba varasematest peat\u00fckkidest oled lugenud, on keemias \u00fcks p\u00f5hilisi suurusi ainehulk, mida kasutatakse erinevates arvutustes. Ainehulga kaudu saame reaktsiooniv\u00f5rrandite p\u00f5hjal arvutada, kui palju ainet reageerib v\u00f5i tekib. Samas tegeldakse keemias\u00a0 v\u00e4ga palju lahustega, mida saame m\u00f5\u00f5ta ruumala\u00fchikutes:\u00a0cm3, dm3, &#8230;<\/p>\n","protected":false},"author":269,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"class_list":["post-10","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/10","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/users\/269"}],"replies":[{"embeddable":true,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/comments?post=10"}],"version-history":[{"count":14,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/10\/revisions"}],"predecessor-version":[{"id":1032,"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/pages\/10\/revisions\/1032"}],"wp:attachment":[{"href":"https:\/\/sisu.ut.ee\/huvitavkeemia\/wp-json\/wp\/v2\/media?parent=10"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}