Answer:

Explanation:
Each value on the scale represents 10 times the previous value.
Subtract 1 from 6, so 10 to the power of 5 is the difference.
The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol
<h3>Further explanation</h3>
Given
Ratio of the concentrations of the products to the concentrations of the reactants is 22.3
Temperature = 37 C = 310 K
ΔG°=-16.7 kJ/mol
Required
the free energy change
Solution
Ratio of the concentration : equilbrium constant = K = 22.3
We can use Gibbs free energy :
ΔG = ΔG°+ RT ln K
R=8.314 .10⁻³ kJ/mol K

Dangerous
hope i helpedXD
Answer:
The volume of carbon dioxide gas generated 468 mL.
Explanation:
The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%
Mass of tablet = 3.45 g
Mass of bicarbonate =
Moles of bicarbonate ion = 

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:
of carbon dioxide gas
Moles of carbon dioxide gas n = 0.01840 mol
Pressure of the carbon dioxide gas = P = 1.00 atm
Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K
Volume of the carbon dioxide gas = V
(ideal gas equation)

1 L = 1000 mL
0.468 L =0.468 × 1000 mL = 468 mL
The volume of carbon dioxide gas generated 468 mL.