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Georgia [21]
3 years ago
12

What is the balanced equation for KOH+CO2—>K2CO3+H2O

Chemistry
1 answer:
insens350 [35]3 years ago
3 0

The balanced equation :

2KOH+CO₂⇒K₂CO₃+H₂O

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

KOH+CO₂⇒K₂CO₃+H₂O

  • give coeffiecient

aKOH+bCO₂⇒K₂CO₃+cH₂O

  • make equation

K, left=a, right=2⇒a=2

H, left=a, right=2c⇒a=2c⇒2=2c⇒c=1

O, left=a+2b, right=3+c⇒a+2b=3+c⇒2+2b=3+1⇒2b=2⇒b=1

the equation becomes :

2KOH+CO₂⇒K₂CO₃+H₂O

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The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?
damaskus [11]
When the reaction equation is:

HF ↔ H+   +   F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X

∴ X = 0.015 M

∴[H+] = X = 0.015

when PH = -㏒[H+]

∴PH = -㏒0.015

        = 1.8
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