Answer:
u meant the answer to this ques
Explanation:
22
×12
------
44
22 ×
----------
264
-----------
or the answer can be 22×12=264
A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).
1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).
2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.
3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.
B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).
1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).
2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.
3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).
4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.
5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).
6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.
Answer:
-30.7 kj/mol
Explanation:
The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula: ∆Go ’= -RTln K’eq
where,
R = -8.315 J / mo
T = 298 K
For reaction,
1. K′eq1=270,
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 270
= - 8.315 x 298 x 5.59
= - 13,851.293 J / mo
= - 13.85 kj/mol
2. K′eq2=890
∆Go ’= -RTln K’eq
= - 8.315 x 298 x ln 890
= - 8.315 x 298 x 6.79
= - 16.82 kj/mol
therefore, total standard free energy
= - 13.85 + (-16.82)
= -30.7 kj/mol
Thus, -30.7 kj/mol is the correct answer.
Solute is something that is being dissolved { ex : sugar , salt}
Solvent is something that has ability to dissolve things { ex : water}
False because the solvent is present in larger amounts...
Answer:


Explanation:
Hello,
Considering the given reaction's stoichiometry, grams of oxygen result:

Moreover, the mass of produced carbon dioxide turns out:

Best regards.