Answer:
- 178 ºC
Explanation:
The ideal gas law states that :
PV = nRT,
where P is the pressure, V is the volume, n is number of moles , R is the gas constant and T is the absolute temperature.
For the initial conditions :
P₁ V₁ = n₁ R T₁ (1)
and for the final conditions:
P₂V₂= n₂ R T₂ where n₂ = n₁/2 then P₂ V₂ = n₁/2 T₂ (2)
Assuming V₂ = V₁ and dividing (2) by Eqn (1) :
P₂ V₂ = n₁/2 R T₂ / ( n₁ R T₁) then P₂ / P₁ = 1/2 T₂ / T₁
4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K
T₂ = 95 - 273 = - 178 º C
Protons/Electrons: 92
Neutrons: 140
***REMEMBER: There is always the same amount of protons and electrons. :)
Moles Chlorine to grams = 35.453 grams.
moles Chlorine to grams = 70.906 grams.
moles Chlorine to grams = 106.359 grams.
moles Chlorine to grams = 141.812 grams.
moles Chlorine to grams = 177.265 grams.
moles Chlorine to grams = 212.718 grams.
<h3>Answer:</h3>
64 g O₂
<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>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Divide/Multiply [Cancel Units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂