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Mariana [72]
2 years ago
10

If you add a solution of NaOH to a solution of H₂CO₃, two reactions occur, one after the other. Write the chemical equations for

these two reactions. (Hint: NaOH dissociate into Na+ and OH-, and the hydroxide ion is the actual base).
Chemistry
1 answer:
Crazy boy [7]2 years ago
6 0

We have a solution of NaOH and H₂CO₃

First, NaOH will dissociate into Na⁺ and OH⁻ ions

The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃

The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule

If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃

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What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

K_{a}=1.8\times 10^{-5}

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Considering the ICE table for the dissociation of acid as:-

\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}

The expression for dissociation constant of acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}

1.8\times 10^{-5}=\frac{x^2}{1.8-x}

1.8\left(1.8-x\right)=100000x^2

Solving for x, we get:

<u>x = 0.00568  M</u>

Percentage ionization = \frac{0.00568}{1.8}\times 100=0.32 \%

<u>Option B is correct.</u>

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The calorific value of a fuel is the amount of heat energy released from the combustion of the fuel under standard conditions.

Dietheylene glycol, isopropanol and paraffin wax are all used as fuels.

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7 0
2 years ago
Find the mass in 6.2 moles of lithium bromide
Art [367]

1
mol of lithium bromide has a mass of
86.85

g
⋅
m
o
l
−
1
. You have
2.6
moles of the stuff.
Explanation:
So mass
=

86.85

g
⋅
m
o
l
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1
×

2.6

m
o
l

=

?
?

g
.
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