Answer:
9.8 × 10²⁴ molecules H₂O
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Organic</u>
<u>Stoichiometry</u>
- Analyzing reaction rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 130 g CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[RxN] 1 mol CH₄ → 2 mol H₂O
[PT] Molar Mass of C: 12.01 g/mol
[PT] Molar Mass of H: 1.01 g/mol
Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O
Answer:
25 grams
Explanation:
You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.
Answer:
806.3g
Explanation:
Given parameters:
Number of moles of silver nitrate = 4.85mol
Unknown:
Mass of silver chromate = ?
Solution:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃
To solve this problem, we work from the known to the unknown;
- The known specie here is AgNO₃ ;
From the balanced chemical equation;
2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄
4.85 moles of AgNO₃ will produce
= 2.43moles of Ag₂CrO₄
- Mass of silver chromate produced;
mass = number of moles x molar mass
Molar mass of Ag₂CrO₄
Atomic mass of Ag = 107.9g/mol
Cr = 52g/mol
O = 16g/mol
Input the parameters and solve;
Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol
So,
Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g
Answer:
B. Ca(OH)2(s)+CO2(aq)-CaCO3(s)+H2O(aq)