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3 years ago

14

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

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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If heat is released from water, what phase change occurs?

gizmo_the_mogwai [7]

Melting phase, the ice is melting into water.

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4 years ago

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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

kogti [31]

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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4 years ago

What is the chemical formula of sulfur dioxide? A. SO B. SO2 C. SO3 D. S2O

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B: SO2
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O=Oxygen
Dioxide if i am correct means 2 oxygen.
so thats one Sulfur and 2 oxygen.

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3 years ago

svlad2 [7]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ It would be a covalent bond most likely. It cannot be an ionic bond as the two elements have the same number of electrons. It also couldn't be a metallic bond as they are obviously not metals.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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4 years ago

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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

KATRIN_1 [288]

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