I think it’d be C. I’m not 100% if it is tho
The one from the cupboard because it is warmer (has more thermal energy) than the one from the refrigerator. When a liquid becomes warmer, the solubility of a gas becomes lower. Therefore, there is already more CO2 in its gaseous state when the soda is warm.
Answer:

Explanation:
They gave us the masses of two reactants and asked us to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 239.27 32.00 207.2
2PbS + 3O₂ ⟶ 2Pb + 2SO₃
m/g: 2.54 1.88
2. Calculate the moles of each reactant

3. Calculate the moles of Pb from each reactant

4. Calculate the mass of Pb

Glucose+oxygen➡️ carbon dioxide+water+ATP
C6H12O6+O2+H2O+ATP.
let me hope that's help full
Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = 
K' =
= 0.11
<em>b. 2NH3(g) ⇌ N2(g) + 3H2(g)</em>
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = 
K' = 
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4