Answer: C
Explanation:
Indicators are organic weak acids or bases with complicated structures.
CH₄ + 2O₂ → CO₂ + 2H₂O
From the equation, we know that methane and carbon dioxide have the same number of moles.
no. of moles of CO₂ produced = no. of moles of methane
= 4.5 × 10⁻³ ÷ (12 + 1×4)
= 2.8125 × 10⁻⁴
∴ mass of CO₂ = 2.8125 × 10⁻⁴ × (12 + 16×2)
= 12.375 × 10⁻³ g
Answer:
your answer will be between 16-17 moles
Explanation:
The combustion of butane is 2 C4H10 + 13 O2 = 8 CO2 + 10 H2O. Every two moles of C4H10 can produce 10 moles of water,
Volume is an extensive physical property and not an intensive one.
Answer:
T₂ = 721 k
Explanation:
Given data:
Initial volume = 285 mL
Initial pressure = 1.88 atm
Initial temperature = 355 K
Final temperature = ?
Final volume = 435 mL
Final pressure = 2.50 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ / P₁V₁
T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL
T₂ = 386062.5 atm. mL. K /535.8 atm. mL
T₂ = 721 k
