Answer:
The volume is 2238,6 L
Explanation:
We use the formula PV=nRT. The conditions STP are: 1 atm of pressure and 273K of temperature:
PV=nRT ---> V=(nRT)/P
V= 100 mol x 0,082 l atm/K mol x 273K/ 1 atm
<em>V=2238, 6L</em>
C: 4 H on the left and right, 4 O on both sides making it balanced
<h3>Answer:</h3>
100 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] 2 mol CH₄
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol CH₄ → 2 mol O₂
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
128 g O₂ ≈ 100 g O₂
The net ionic equation for the neutralization reaction involving equal molar amount amount of HNo3 and KoH is
H^+ + OH^- = H2O (l)
explanation
write the chemical equation
HNO3 (aq) + KOH(aq) = KNO3(aq) +H2O (l)
ionic eequation
H^+(aq) + NO3^- (aq) + K^+9aq) OH^-(aq) = K^+ (aq) + NO3^-(aq) + H2O(l)
cancel the spectator ions( ions which does not take place in equation ) for this case is NO3^- and No3^-
thus the net ionic is
H^+(aq) + OH^- (aq) = H2O(l)
As the independent variable changes, the dependent variable never changes.