The answer would be products
The answer is 1.56L. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles (or a number of particles) of gas when the temperature and pressure are held constant.
V∝n
V₁/n₁m= V₂/n₂
V₁ = initial volume of gas = 12.5 L
V₂ = final volume of gas = ?
n₁ = initial moles of gas = 0.016 mole
n₂ = final moles of gas = 0.016-0.007 = 0.002 mole
V₁/n₁m= V₂/n₂
V₂= 1.56L
Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.
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Answer:
2.41065 grams
Explanation:
Here we have to apply molarity, particularly in reference to the equation molarity = moles of solute / volume. I would like to rewrite this formula, but with respect to the units - grams = moles / Liters,
We can use molarity to determine the number of moles. After doing so, we can determine the mass of the solute with respect to the formula moles = mass / molar mass. The molar mass of NaCl is 58.44 grams.
_______________________________________________________
275 mL = 0.275 L,
Number of Moles of NaCl = 0.150 * 0.275 = 0.04125 moles,
Mass = 0.04125 * 58.44 = 2.41065 grams,
Solution - Mass of NaCl = 2.41065 grams
<u><em>Hope that helps!</em></u>
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
Maybe if you put it in english i’ll answer.. nahh jp i’m fluent