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dimulka [17.4K]
3 years ago
9

The solubility of agcl(s) in water at 25 ∘c is 1.33×10−5mol/l and its δh∘ of solution is 65.7 kj/mol. what is the solubility at

47.7 c
Chemistry
1 answer:
slamgirl [31]3 years ago
8 0
According to this equation:

AgCl(s) ↔ Ag+(aq)  + Cl-(aq)

so K1 = [Ag+][Cl-]

when [Ag+] = [Cl-]  we can assume both = X 

and when we have X the solubility = 1.33 x 10^-5 mol / L

by substitution:

∴ K1 = X^2

       = (1.33 x 10^-5)^2

       = 1.77 x 10^-10

by using vant's Hoff equation:

ln(K2/K1) = (ΔH/R)*(1/T2-1/T1)

when ΔH = 65700 J / mol

R = 8.314 

T1 = 25+273 = 298 K

T2 = 47.7 +273 =320.7

by substitution:

∴㏑(K2/1.77 x 10^-10) = (65700/8.314) * ( 1/320.7 - 1/ 298)

by solving for K2 

∴K2 = 2.7 x 10^-11

and when K2 = X^2

∴ the solubility X = √(2.7 x 10^-11)

                             = 5.2 x 10^-6 mol/L  
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Answer:

162.7miles/hr

Explanation:

Given parameters:

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It is mathematically expressed as;

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So, input parameters and solve;

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8 0
3 years ago
Which statement best describes the amount of catalyst that remains at the end of a reaction?
Ivenika [448]
I believe the correct answer is C. The amount of catalyst is the same at the end as at the beginning of the reaction. Catalysts can't be consumed by the reaction thus is not D.
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3 years ago
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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

3 0
3 years ago
What reaction will take place if h2o is added to a mixture of nanh2/nh3? draw the products of the reaction (without sodium ion)?
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The chemical reaction is given by:

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Now, when ammonia i.e.NH_{3} reacts with water results in the formation of ammonium hydroxide i.e. NH_{4}OH

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Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.


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