Answer:
80L
Explanation:
V1/T1 = V2/T2
V2 = V1 T2/T1
T1 = 300K
V1 = 60L
T2 = 400K
V2 = ?
V2 = V1 T2/T1
V2 = (60L)(400K) / (300K)
V2 = 80L
Answer:
2:a-heterogenous
b-homogenous
c-heterogenous
d-heterogenous with water
3:Filtration is used to separate insoluble particles from a solution
Distillation is used to separate liquids with close but different boiling points e.g water and ethanol
Explanation:
2: homogenous mixtures form a uniform layer meaning that a mixture containing more than one layer is heterogenous
The molarity of NaOH needed is calculated as follows
calculate the moles of KhC8h4O4
that is moles = mass/molar mass of KhC8h4O4(204.22 g/mol)
=0.5632g /204.22g/mol= 2.76 x10^-3 moles
write the equation for reaction
khc8h4O4 + NaOH ---> KNaC8h4O4 + H2O
from the equation above the reacting ratio of KhC8h4O4 to NaOh is 1:1 therefore the moles of Naoh is also 2.76 x10^-3 moles
molarity of NaOh = (moles of NaOh / volume ) x 1000
that is { (2.76 x10^-3) / 23.64} x100 =0.117 M
As the gas cools it condenses and becomes a liquid its atoms also become smaller
Answer:
1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Explanation:
1) Possible reactions
2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Mass of each metal
a) Mass of Cu
The waste was the unreacted copper.
Mass of Cu = 2.5 g
b) Masses of Al and Fe
We have two relations
:
Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g
H₂ from Al + H₂ from Fe = 6.38 L at NTP
i) Calculate the moles of H₂
NTP is 20 °C and 1 atm.

(ii) Solve the relationship
Let x = mass of Al. Then
7.5 - x = mass of Fe
Moles of Al = x/27
Moles of Fe = (7.5 - x)/56
Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x
/18
Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56
∴ x/18 + (7.5 - x)/56 = 0.2652
56x + 18(7.5 - x) = 267.3
56x + 135 - 18x = 267.3
38x = 132.3
x = 3.5 g
Mass of Al = 3.5 g
Mass of Fe = 7.5 g - 3.5 g = 4.0 g
The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g