Answer:
The number of moles of CaCO3 on the bag is 112.90 moles
Explanation:
number mole (n) = mass (m) divided by molecular mass (Mm)
Mm of CaCO3 = 100.0869 g/mole
mass in grams = 11.3 Kg x (10^3 g/1 Kg) = 11300 grams
number of moles (n) = 11300 grams divided by 100.0869 grams per mole = 112.90 moles of CaCO3 in the bag.
Answer:
<h3>The answer is 43.89 mL</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula

From the question
mass = 57.5 g
density = 1.31 g/mL
We have

We have the final answer as
<h3>43.89 mL</h3>
Hope this helps you
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C