Okay so your answer for n is
1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
Answer:

Explanation:
Hello!
In this case, according to the following chemical reaction:

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

Best regards!
Answer:
Here's what I get
Explanation:
CH₃CH₂CH₂CH₂CH₂CH₃ — hexane
CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted
CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted
CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane
CH₃CH₂CHCICH₂CH₃ — 3-chloropentane
<u>Answer:</u> The
for the reaction is 51.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of carbon and water follows:

The intermediate balanced chemical reaction are:
(1)
( × 2)
(2)
( × 2)
(3)

The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B2%5Ctimes%20%5CDelta%20H_1%5D%2B%5B2%5Ctimes%20%5CDelta%20H_2%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-393.7%29%29%2B%282%5Ctimes%20%28-285.9%29%29%2B%281%5Ctimes%20-%28-1411%29%29%5D%3D51.8kJ)
Hence, the
for the reaction is 51.8 kJ.