A GAS SAMPLE HAS A TEMPERATURE OF 18 CELSIUS WITH AN UNKNOWN VOLUME. THE SAME GAS HAS A VOLUME OF 436 ML WHEN THE TEMPERATURE IS 91 CELSIUS WITH NO CHANGE IN THE PRESSURE AND AMOUNT OF GAS
WHAT WAS THE INITIAL VOLUME IN MILLILITERS OF THE GAS

349 mL

Explanation:

Given data

If we consider the has as an ideal gas, we can calculate the initial volume of the gas sample using the Charles' law.

V₁ / T₁ = V₂ / T₂

V₁ = V₂ × T₁ / T₂

V₁ = 436 mL × 291 K / 364 K = 349 mL

option c: 1605.1 kj

the balanced chemical reaction for combustion of methane is as follows:

here, enthalpy formation of is -74.7 kj/mol, 0 kj/mol, -393.5 kj/mol and -241.82 kj/mol respectively.

the heat released or can be calculated as follow:

putting the values,

thus, heat released by combustion of 1 mol of methane is 802.54 kj/mol. thus, for 2 mol, heat released will be:

therefore, heat released by the combustion of 2 mol of methane is 1605.1 kj.