8) 21.504 liters of gas
9) 122.5 g
10) 6.022*10^23
Explanation:
8)
1 mole= 22.4 liters
to calculate the volume of gas of 0.960 moles of CH4
22.4*0.960/1
= 21.504 liters of gas
9)
2.0 mole*18.02 g/ 1 mole
=122.5 g
10)
2.0 l * 1 mole/22.4 l
6.022*10^23
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).
A chemical substance has the characteristics that it cannot be separated by physical methods. Seawater and milk can be separated by sedimentation, and air has different components depending on other aspects (such as elevation). Only ammonia is a substance. (thus it can have a formula: NH<span>3)</span>
<span>The student is incorrect because helium has 2 valence electrons and it's in group 18 because the first energy level is full. Although helium is placed in Group 18 which generally has 8 valence electrons, it does not have 8 valence electrons as the student suggested. It was grouped together with the noble gases because it exhibits similar properties with them. </span>
