<u>Answer:</u> 2.00 atm
<u>Explanation:</u>
The gas is kept under the same temperature in this problem. Assuming the amount of gas is constant, we can apply the Boyle's law.
The Boyle's law equation,
P₁V₁ = P₂V ₂
Plug in the values,
1.00 atm x 4.0 L = P₂ x 2.0 L
Simplify,
4.00 atm L = 2 P₂ L
Now flip the equation,
2 P₂ L = 4.00 atm L
Dividing both sides by 2 we get,
P₂ = 2.00 atm
The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
Learn more: brainly.com/question/13164491
Carbon discovered in Prehistoric times.. Discoverer will probably never be known
This would be the molar mass.
Increase in temperature makes the atoms speed up, and decrease in temperature makes the atoms move slower.
