Answer:
0.56 g
Explanation:
<em>A chemist determines by measurements that 0.020 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.</em>
Step 1: Given data
Moles of nitrogen gas (n): 0.020 mol
Step 2: Calculate the molar mass (M) of nitrogen gas
Molecular nitrogen is a gas formed by diatomic molecules, whose chemical formula is N₂. Its molar mass is:
M(N₂) = 2 × M(N) = 2 × 14.01 g/mol = 28.02 g/mol
Step 3: Calculate the mass (m) corresponding to 0 0.020 moles of nitrogen gas
We will use the following expression.
m = n × M
m = 0.020 mol × 28.02 g/mol
m = 0.56 g
Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry ) 
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
Answer:
c = 0.13 j/ g.°C
Explanation:
Given data:
Mass of mercury = 29.5 g
Initial temperature = 32°C
Final temperature = 161°C
Heat absorbed = 499.2 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 161°C - 32°C
ΔT = 129 °C
Q = m.c. ΔT
c = Q / m. ΔT
c = 499.2 j / 29.5 g. 129 °C
c = 499.2 j / 3805.5 g. °C
c = 0.13 j/ g.°C
