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Vlad1618 [11]
3 years ago
14

CH4(g) + 2O2(g) = CO2(g) + 2H2O(g)

Chemistry
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

0.04 moles of CH₄  are required

Explanation:

Given data:

Number of moles of CH₄ required = ?

Number of moles of O₂ present = 0.08 mol

Solution:

Chemical equation:

CH₄ + 2O₂     →   CO₂ + 2H₂O      

now we will compare the moles of oxygen and methane.

                      O₂          :        CH₄

                       2           :           1

                     0.08        :        1/2×0.08 = 0.04

Thus, 0.04 moles of CH₄  are required.

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Enter your answer in the provided box. Sodium stearate (C17H35COONa) is a major component of bar soap. The Ka of the stearic aci
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Answer:

11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.

Explanation:

Concentration of sodium stearate acid : c

Moles of sodium stearate = \frac{3.96 g}{306 g/mol}=0.01294 mol

Volume of the solution = 10.0 mL = 0.010 L

c=\frac{0.01294 mol}{0.010 L}=1.294 M

C_{17}H_{35}COONa\rightleftharpoons C_{17}H_{35}COO^-+Na^+

[C_{17}H_{35}COO^-]=c=1.294 M

C_{17}H_{35}COO^-+H_2O\rightleftharpoons C_{17}H_{35}COOH +OH^-

initially c

c           0    0

At equilibrium

(c-x)       x    x

Dissociation constant of an acid = K_a=1.3\times 10^{-5}

Expression of a dissociation constant of an acid is given by:  

K_a=\frac{[C_{17}H_{35}COOH][OH^-]}{[C_{17}H_{35}COO^-]}

K_a=\frac{(x)^2\times x}{(c -x)}

1.3\times 10^{-5}=\frac{x^2}{1.294-x}

Solving for x;

x = 0.0041 M

[OH^-]=0.0041 M

The pOH of the solution:

pOH=-\log[OH^-]=-\log[0.0041 M]=2.39

pH = 14 -pOH

pH = 14 - 2.39 = 11.61

11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.

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If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH
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Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)

M_{KOH}=0.113 M\\V_{KOH}=79.06 mL

V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????

M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}

M_{CH3COOH} = \frac{0.113*79.06}{95.27}

M_{CH3COOH} = 0.094 M

Moles of CH_3COOH = 95.27* 10^{-3}* 0.094

=0.0090 moles

Moles of  KOH = 79.06*10^{-3}*0.113

= 0.0090 moles

The equation for the reaction can be expressed as :

CH_3COOH     +      KOH     ----->      CH_3COO^{-}K^+      +     H_2O

Concentration of CH_3COO^{- ion = \frac{0.0090}{Total volume (L)}

= \frac{0.0090}{(95.27+79.06)} *1000

= 0.052 M

Hydrolysis of  CH_3COO^{- ion:

CH_3COO^{-      +       H_2O      ----->      CH_3COOH       +     OH^-

K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}

⇒    \frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}

=     0.5494*10^{-9}= \frac{x*x}{0.052-x}

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

0.5494*10^{-9}= \frac{x^2}{0.052}

x^2 = 0.5494*10^{-9}*0.052

x^2 = 0.286*10^{-10

x = \sqrt{0.286*10^{-10

x =0.535*10^{-5}M

[OH] = x =0.535*10^{-5}

pOH = -log[OH^-]

pOH = -log[0.535*10^{-5}]

pOH = 5.27

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0
3 years ago
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