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pochemuha
4 years ago
9

When looking at the equilibrium between silver bromide and its aqueous ions, what could be added to solution to promote precipit

ation of silver bromide?
Chemistry
1 answer:
jeka57 [31]4 years ago
5 0

Answer:

NaBr

Explanation:

When AgBr is dissolved in water, the following equilibrium is set up in solution;

AgBr(s)⇄Ag^+(aq) + Br^- (aq)

If we dissolve NaBr in the water, a common ion (Br^-) is now introduced into the system. This increases the concentration of Br^- and favours the reverse reaction hence more AgBr is precipitated. This is known as common ion effect.

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An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 15.71 g CaCl2. After performing the experim
OlgaM077 [116]

Answer:

93.15 %

Explanation:

We have to start with the chemical reaction:

CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl

Now, we can balance the reaction:

CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl

Our initial data are the 15.71 g of CaCl_2, so we have to do the following steps:

1) <u>Convert from grams to moles of CaCl_2 using the molar mass (110.98 g/mol).</u>

2) <u>Convert from moles of CaCl_2 to moles of CaCO_3 using the molar ratio. ( 1 mol CaCl_2= 1 mol of CaCO_3).</u>

3) <u>Convert from moles of CaCO_3 to grams of CaCO_3 using the molar mass. (100 g/mol).</u>

15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3

Finally, we can calculate the yield percent:

%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%

I hope it helps!

5 0
4 years ago
When water is heated in an ocean the liquid water changes from and turns into?
sasho [114]
Hello!

The answer is Evaporation

When you add heat to water, it boils, releasing gas into the air. This is called Evaporation. The water changes from a liquid to a gas.

Hope this helped!
6 0
3 years ago
0.7 moles of uranium to grams of uranium​
alexandr1967 [171]

Answer: 166.6g

Explanation: 0.7mol x 238g/1mol

8 0
3 years ago
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.
tatiyna

Answer:

[H_2]_{eq}=0.183M

[I_2]_{eq}=0.183M

[HI]_{eq}=0.025M

Explanation:

Hello.

In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

2HI\rightleftharpoons H_2+I_2

Whereas the law of mass action (equilibrium expression) is:

Kc=\frac{[H_2][I_2]}{[HI]^2}

That in terms of initial concentrations and reaction extent or change x turns out:

Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}

And the solution via solver or quadratic equation is:

x_1=0.183M\\\\x_2=0.210M

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:

[H_2]_{eq}=0.183M

[I_2]_{eq}=0.183M

[HI]_{eq}=0.391M-2*0.183M=0.025M

Regards.

6 0
3 years ago
A fluorine ion has 9 protons and a charge of negative one how many electrons are in an ion of fluorine
Ulleksa [173]
-1 charge = the atom has gained one electron

9+1=10
6 0
4 years ago
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