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pishuonlain [190]

4 years ago

9

There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc

acetate, Zn(CH3COO)2, is 43.0 g/L, what is the solubility product Ksp of this compound?

1 answer:

zhuklara [117]4 years ago

8 0

Answer:

$K_{sp}$ of $Zn(CH_{3}COO)_{2}$ is 0.0513

Explanation:

Solubility equilibrium of $Zn(CH_{3}COO)_{2}$ :

$Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}$

Solubility product of $Zn(CH_{3}COO)_{2}$ ( $K_{sp}$ ) is written as- $K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}$

Where $[Zn^{2+}]$ and $[CH_{3}COO^{-}]$ represents equilibrium concentration (in molarity) of $Zn^{2+}$ and $CH_{3}COO^{-}$ respectively.

Molar mass of $Zn(CH_{3}COO)_{2}$ = 183.48 g/mol

So, solubility of $Zn(CH_{3}COO)_{2}$ = $\frac{43.0}{183.48}M$ = 0.234M

1 mol of $Zn(CH_{3}COO)_{2}$ gives 1 mol of $Zn^{2+}$ and 2 moles of $CH_{3}COO^{-}$ upon dissociation.

so, $[Zn^{2+}]$ = 0.234 M and $[CH_{3}COO^{-}]$ = $(2\times 0.234)M=0.468M$

so, $K_{sp}=(0.234)\times (0.468)^{2}=0.0513$

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Magnesium phosphate is used as a dental polishing agent. how many moles of ions (cations and anions together) are in 1 mole of m

S_A_V [24]

The answer to this question would be:5 mol

To answer this question, you need to know the chemical equation of mg3(po4)2 when forming an ion. The equation should be: <span>mg3(po4)2= 3Mg(2+) +2PO4(3-)

That means for 1 mol of </span><span>mg3(po4)2 there will be 3 magnesium(cation) ion and 2 phosphates(anion) ion, the result would be 5 mol of ion.</span>

7 0

4 years ago

At 298 K, Kc = 1.45 for the following reaction 2 BrCl (g) Br2(g) + Cl2(g) A reaction mixture was prepared with the following ini

tester [92]

Answer:

[BrCl]=0.02934M

[Br2]=[Cl2]=0.03533M

Explanation:

First, consider the Kc definition:

$1.5=K_{c}=\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}$

It is necessary to define a new variable, 'x', as the amount of moles of Br2 that are produced. If 'x' moles of Br2 are produced, the moles of the compounds will be calculated as;

$n_{Br_{2}}=n_{{Br_{2}}^{0}} +x\\n_{Cl_{2}}=n_{{Cl_{2}}^{0}} +x\\n_{BrCl}=n_{{BrCl}}^{0}} -2x\\$

Where the zero superscript means the initial moles.

Dividing the last equations by the volume, which is constant because the reaction does not change the total moles number (2 moles of BrCl produce 2 moles, one of Cl2 and another of Br2), we have the molarity equations for all species:

$M_{Br_{2}}=M_{{Br_{2}}^{0}} +x\\M_{Cl_{2}}=\\M_{BrCl}=M_{{BrCl}}^{0}} -2x\\$

And now 'x' is a change in molarity.

Replacing these in the Kc equation we have:

$1.45=\frac{({M_{{Br_{2}}}^{0}}+x)({M_{{Cl_{2}}}^{0}}+x)}{({M_{{BrCl}}^{0}}-2x)^{2}}$

Where the only unknown is 'x'. So, let's solve the equation:

$1.45=\frac{(0.03+x)(0.03+x)}{(0.04-2x)^{2} } \\1.45(0.04-2x)^2=(0.03+x)^2\\1.45(0.0016-0.16x+4x^2)=0.0009+0.06x+x^2\\4.8x^2-0.292x+0.00142=0\\x_{1}=0.0555\\x_{2}=0.00533$

The result $x_{1}=0.0555$ lacks of sense because it will give a negative concentration for BrCl, so the result is $x_{2}=0.00533$ <u>.</u>

Applying the result, the concentrations at equilibrium are:

$M_{Br_{2}}=M_{Cl_{2}}=0.03+0.00533=0.03533M\\M_{BrCl}=0.04-2*0.00533=0.02934M$

If you calculate Kc with this concentrations it will give 1.45 as a result.

Greets, I will be happy to solve any doubt you have.

8 0

4 years ago

Why is the "coming up" list on the dashboard useful?

kirill115 [55]

I'm teaching with Canvas. On my dashboard, under the Coming Up section, none of the assignments that are on the calendar (populated b/c they are "assignments") are displaying. The next one is due in 36 hours so I think it should display and it is listed on the calendar.

6 0

3 years ago

Can someone please help me please?

OleMash [197]

I would say the second option

Hope this helps *smiles*

8 0

3 years ago

Avogadro’s law states that the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both

natima [27]

Answer: $\frac{V_1}{V_2}=\frac{n_1}{n_2}$

Explanation:

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

$n_1$ = initial moles of gas

$n_2$ = final moles of gas

3 0

3 years ago