Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
Explanation:
From the question we are told that:
Heat Capacity 
Mass of water 
Initial Temperature of Aluminium 
Initial Temperature of Water 
Final Temperature of Water 
Generally
Heat loss=Heat Gain
Therefore
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
And the significant amount of volume can be differed by its solitude
Answer:
4.5 moles of lithium sulfate are produced.
Explanation:
Given data:
Number of moles of lead sulfate = 2.25 mol
Number of moles of lithium nitrate = 9.62 mol
Number of moles of lithium sulfate = ?
Solution:
Chemical equation:
Pb(SO₄)₂ + 4LiNO₃ → Pb(NO₃)₄ + 2Li₂SO₄
Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.
Pb(SO₄)₂ : Li₂SO₄
1 : 2
2.25 : 2/1×2.25 = 4.5 mol
LiNO₃ : Li₂SO₄
4 : 2
9.62 : 2/4×9.62 = 4.81 mol
Pb(SO₄)₂ produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of Li₂SO₄.
