<h3>
Answer:</h3>
5.58 × 10²⁴ molecules 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>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
9.27 mol O₂ (Oxygen)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules. We are given 3 sig figs.</em>
5.58239 × 10²⁴ molecules O₂ ≈ 5.58 × 10²⁴ molecules O₂
In the space of 1 liter, methane gas (ch4) is reacted with 6 moles of water vapor according to the reaction: if at equilibrium state is obtained 4 moles of hydrogen gas, how many moles of methane gas is needed for the equilibrium reaction?
the reaction is
CH4(g) + 2H2O(g) ----> CO2(g) + 4H2 (g)
Kc = 16 / 3
Kc = [CO2] [H2]^4 / [CH4] [H2O]^2
given :
equilibrium concentration
[H2] = 4 moles
so equilibrium concentration of CO2 must be 1 mole
equilibrium concentration of H2O = 6 - 2 = 4
putting values
16 /3 = [1] [4]^4 / [CH4] [4]^2
[CH4] = 0.333 moles
so moles of CH4 required = 1.33 moles
A. The products have a lower potential energy than the reactants.
Answer:
Option C:- that is equal to mass of an proton.
Explanation:
Protons and neutrons have approximately the same mass, about 1.67 × 10-24 grams, which scientists define as one atomic mass unit (amu) or one Dalton. While electron has mass of 9.31 ×10⁻¹⁹.
Answer:
The rate at which ammonia is being produced is 0.41 kg/sec.
Explanation:
Haber reaction
Volume of dinitrogen consumed in a second = 505 L
Temperature at which reaction is carried out,T= 172°C = 445.15 K
Pressure at which reaction is carried out, P = 0.88 atm
Let the moles of dinitrogen be n.
Using an Ideal gas equation:
According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.
Then 12.1597 mol of dinitrogen will produce :
of ammonia
Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol
=413.43 g=0.41343 kg ≈ 0.41 kg
505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.
