Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
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Answer:
Mole fraction of Nacl is 0.173
Explanation:
we know that
where,
P
sol - the vapor pressure of the solution
χ solvent - the mole fraction of the solvent
P
∘
solvent - the vapor pressure of the pure solvent
This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at
25
°
C You can use an online calculator to find that the vapor pressure of pure water at 25 C is equal to about 23.8 torr
.
=0.827
Also we know that
This means that the mole fraction of sodium chloride is
χ_{Nacl}= 1-Χ_{water}
= 1-0.827 =0.173
21.69mL - 20.70mL = .99mL Fe
7.8 g / .99 mL = 7.9g/mL
you have to show us the rest of it because we have no idea what your looking at. I'm sorry
