Answer:
a. 0.171M
b. 0.0938M
c. 0.284
d. 1.99atm
e. 1.88
Explanation:
Hello,
In this case, for the given reaction whose balance should be corrected as:
For which the law of mass action, in terms of the change due to stoichiometry and the reaction extent, turns out:
Thus, the initial concentration of hydrogen sulfide is:
Now, since the equilibrium amount of sulfur is given, the change due to equilibrium reaching is:
Therefore:
a. Equilibrium concentration of hydrogen:
b. Equilibrium concentration of hydrogen sulfide:
c.) Equilibrium constant, Kc:
d.) Partial pressure of sulfur gas:
e. Kc, for the reaction:
In that case, it equals the inverse halved initial reaction, whose modification is related as:
Best regards.
Answer:
While carbon dioxide is in the air waiting to be reabsorbed it traps the sun's heat.
Answer:
Explanation:
<u>1. Oxidation half-reaction (given)</u>
- 2S₂O₃²⁻ (aq) → S₄O₆²⁻ + 2e⁻
<u>2. Reduction half-reaction:</u>
Reduction is the gain of electrons with the consequent reduction in the number of oxidation.
The starting reactant is iodine which is diatomic; thus, it is I₂.
Each iodine atom gains one electron; thus, in total 2 electrons are gained and each I atom in solution will become an I⁻ anion.
Half-reaction:
- I₂ (aq) + 2e⁻ → 2I⁻(aq) ↔ answer
The overall reaction of thiosulfate anion with iodine is obtained when you add the two half-reactions:
- 2S₂O₃²⁻ (aq) → S₄O₆²⁻ + 2e⁻
===============================
(the electrons are canceled)
- 2S₂O₃²⁻ (aq) + I₂ (aq) → S₄O₆²⁻ + 2I⁻(aq)
Explanation:
A balanced redox chemical equation is an equation that contains same number of atoms on both reactant and product side. Also, it contains same total charge on reactant and product side.
For example,
Number of atoms on reactant side are as follows.
Mn = 1
O = 4
Ag = 1
H = 1
Number of atoms on product side are as follows.
Mn = 1
O = 1
Ag = 1
H = 2
Therefore, to balance the equation multiply on reactant side by 8 and multiply on product side by 4.
Now, total charge on reactant side is +7 but total charge on product side is +2. Therefore, to balance the charges we multiply Ag(s) on reactant side by 5 and on product side by 5.
Therefore, completely balanced redox reaction equation will be as follows.
<h3>
Answer:</h3>
0.0253 mol H₂O
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 0.456 g H₂O (water)
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.025305 mol H₂O ≈ 0.0253 mol H₂O