Answer:
29.98kg
Explanation:
12.0 gallons * (3.78541178 liters/gallon) * (1000 mL/liter) * (0.66 g/mL) * (1 kg/1000 g) = 29.98 kg
None of the above.
1 mole filled with gas at STP occupies
=22.4 L
∴ 3mole of kr gas at STP occupies
= 3 × 22.4
= 67.2 L
Answer:

Explanation:
Hello!
In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

Best regards!
<span>The half-life of a first-order reaction is determined as follows:
</span>t½<span>=ln2/k
From the equation, we can calculate the </span><span>first-order rate constant:
</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³
When we know the value of k we can then calculate concentration with the equation:
A₀ = 2 g/100 mL
t = 2.5 h = 150min
A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml
= 6.3 × 10⁻⁴ mg / 100ml