Help!!!!!!please it’s due tonight
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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>
- Naming carbons
<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
We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.
The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.
For any gaseous substance, density of gas is directly proportional to pressure of gas.
This can be explained from idial gas edquation:
PV=nRT
PV=RT [where, w= mass of substance, M=molar mass of substance]
PM=RT
PM=dRT [where, d=density of thesubstance]
So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.
Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.