Answer:
1.3 M/s
Explanation:
Step 1: Given data
- Initial amount of methane (ni): 8.0 mol
- Final amount of methane (nf): 0 mol (it reacts completely)
- Volume of the container (V): 2.00 L
Step 2: Calculate the average rate of consumption of methane
Methane burns in excess oxygen according to the following equation.}
CH₄ + 2 O₂ ⇒ CO₂ + 2 H₂O
We can calculate the average rate of consumption of methane (r) using the following expression.
r = - Δn/V × t = - (nf- ni) / V × t
r = - (0 mol - 8.0 mol) / 2.00 L × 3.2 s = 1.3 M/s
Answer:
8.44 atm
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2.25 L
Initial temperature (T₁) = 350 K
Initial pressure (P₁) = 1.75 atm
Final volume (V₂) = 1 L
Final temperature (T₂) = 750 K
Final pressure (P₂) =?
The final pressure of the gas can be obtained as illustrated below:
P₁V₁/T₁ = P₂V₂/T₂
1.75 × 2.25 / 350 = P₂ × 1 / 750
3.9375 / 350 = P₂ / 750
Cross multiply
350 × P₂ = 3.9375 × 750
350 × P₂ = 2953.125
Divide both side by 350
P₂ = 2953.125 / 350
P₂ = 8.44 atm
Thus, the final pressure of the gas is 8.44 atm.
Answer: OA is a balanced equation.
Explanation: For an equation to be balanced, the number of atoms and the total charge of each element in the reaction, of the product and reactant side must be the same.
Considering OA; 6 atoms of nitrogen is present in the reactant side. Shifting to the product side, 6 atoms of nitrogen is also present. Taking the other element in the reaction into account, the number of atoms of each element balances on both sides.
