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3 years ago

7

Which of the salts on the Slightly Soluble Salts tab has the greatest solubility in water based on moles of ionic compound disso

lved per liter of solution? mercury(II) bromide thallium(II) sulfide

1 answer:

lesantik [10]3 years ago

7 0

Answer:

mercury (II) Bromide

Explanation:

A small value of Ksp indicates a insoluble substance.

the solubility of mercury(II) bromide is 6.2 x 10-20

while that of thallium(II) sulfide is

6.0 x 10-22

from these values thallium sulfide has the smallest solubility which means it is highly insoluble in per liters of water.

while mercury (II) Bromide is more soluble.

there the answer is mercury (II) Bromide

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Which of these solutes raises the boiling point of water the most?

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(E) ionic aluminum fluoride (AlF3)

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3 years ago

according to the balanced chemical equation below,how many grams of h20 are produced if 3.98g of co2 were produced. 2C18H18 + 25

telo118 [61]

Answer:

Primero debes usar los gramos de co2 y luego buscar su peso molecular, luego de eso usar la relación de moles entre CO2 y H2O y por último buscar el pm del H2O pata ver cuantos gramos de produce.

Explanation:

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In the reaction: NH4+ + H2O ⮀ NH3 + H3O+ water is acting as a(n)

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2 years ago

Read 2 more answers

Using the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluor

jenyasd209 [6]

Answer:

27.9 g

Explanation:

CsF + XeF₆ → CsXeF₇

First we <u>convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles</u>, using its<em> molar mass</em>:

Molar mass of CsXeF₇ = 397.193 g/mol

73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇

As <em>1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇</em>, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.

Now we <u>convert 0.184 moles of CsF to moles</u>, using the <em>molar mass of CsF</em>:

Molar mass of CsF = 151.9 g/mol

0.184 mol * 151.9 g/mol = 27.9 g

4 0

2 years ago

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

5 0

3 years ago