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:
180 mg
Explanation:
For a first-order reaction, we can calculate the amount of aspirine (A) at a certain time (t) using the following expression.
where,
k: rate constant
A₀: initial amount
If we know the half-life (
) we can calculate the rate constant.
When t = 4 h and A₀ = 400 mg, A is:
Use a ratio to find out that x/190.2 = 74.5/100
hope this helps
Answer:
A. Intramolecular interactions are generally stronger.
B. a. Only intermolecular interactions are broken when a liquid is converted to a gas.
Explanation:
<em>A. Which is generally stronger, intermolecular interactions or intramolecular interactions?</em>
Intramolecular interactions, in which electrons are gained, lost or shared, constitute true bonds and are one or two orders of magnitude stronger than intermolecular interactions.
<em>B. Which of these kinds of interactions are broken when a liquid is converted to a gas?</em>
When a liquid vaporizes, the intermolecular attractions are broken, that is, molecules get more separated. However, true bonds are not broken which is why the molecules keep their chemical identity.
