CH4(9) + H2O(g) <> CO(g) + 3 H2(g)
1 answer:
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Answer:
5.702 mol K₂SO₄
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Compounds
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 993.6 g K₂SO₄
[Solve] moles K₂SO₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of K: 39.10 g/mol
[PT] Molar Mass of S: 32.07 g/mol
[PT] Molar mass of O: 16.00 g/mol
Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄
The question is incomplete, here is the complete question.
If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.
a. 28.8 b. 30.9 c. 35.4 d. 32.2
Answer : The half-life of strontium-90 is 28.8 years.
Explanation :
This is a type of radioactive decay and all radioactive decays follow first order kinetics.
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by :
where,
k = rate constant
t = time taken for decay process = 9.00 year
a = initial amount or moles of the reactant = 1.000 g
a - x = amount or moles left after decay process = 0.805 g
Putting values in above equation, we get:
To calculate the half-life, we use the formula :
Therefore, the half-life of strontium-90 is 28.8 years.