The formula is m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
1 mole = 6.022 E23 atoms
0.97 mole=?
6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
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
Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
