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

The balanced equation :

2KOH+CO₂⇒K₂CO₃+H₂O

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

aKOH+bCO₂⇒K₂CO₃+cH₂O

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

i think that the answer is c

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally