The answer is the first one
Where are the equations ..
Sorry....
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).
Answer:
Option (A) the solid X is ground to a fine powder.
Explanation:
X(s) + 2B(aq) → X+(aq) + B2(g)
In the reaction above, the rate of the reaction will be highest, when X being a solid is ground to fine powder.
Grounding X to fine powder simply means increasing the surface area of X.
An increase in surface area of reactants will definitely increase the rate of reaction because the particles of the solid will collide with the right orientation and hence speed up the reaction rate.
