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KiRa [710]
3 years ago
6

Is this correct im so confused.

Chemistry
1 answer:
inysia [295]3 years ago
4 0

Answer:

<em>the last option!!!!</em>

Explanation:

because its right

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Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M
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The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>
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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]

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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}

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4 years ago
A student calculated the standard enthalpy change of formation of ethane C2H6 using a method based on standard enthalpy change o
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C2H6 + 302 -------> 2C + 3H20
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