Answer:
45200J
Explanation:
Given parameters:
Heat of vaporization of water = 2260J/g
Mass of steam = 20g
Temperature = 100°C
Unknown:
Energy released during the condensation = ?
Solution:
This change is a phase change and there is no change in temperature
To find the amount of heat released;
H = mL
m is the mass
L is the latent heat of vaporization
Insert the parameters and solve;
H = 20g x 2260J/g
H = 45200J
Answer:
aluminum is highly reactive
This is the three cases that help to determine the minimum concentration of KOH required for precipitation
Part a) 1.5×10^−2 M K CaCl2
Part b) 2.3×10^−3 M Fe (NO3)2
Part c) 2.0×10^−3 M MgBr2
a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^-
ksp = 1.5*10^-2 + x^2
4.68*10^-6 = 1.5*10^-2 + x^2
x= [KOH] = 0.01766
b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3
Fe (OH)2 <=> Fe^2+ + 2OH^-
ksp = 2.3*10^-3 + x^2
4.87*10^-17 = 2.3*10^-3 + x^2
x= 1.46*10^-7
c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr
Mg (OH) 2 <=> Mg^2+ + 2OH^-
ksp = 2.0*10^-3 + x^2
2.06*10^-13 = 2.0*10^-3 + x^2
x= 1.015*10^-5
Answer:
The mass of oxygen needed to be pumped in the aquarium is 10 g
Explanation:
Here the value of the degree of solubility of oxygen in water is sought
The solubility of oxygen in fresh water is approximately 1.22 × 10⁻³ mol·dm⁻³ which is equivalent to approximately 40 mg/L
The volume of water the aquarium holds = 250 L
The mass of oxygen required = The solubility of oxygen in water × (The volume of water in the aquarium)
Therefore, the mass of oxygen needed to be pumped in the aquarium = 40 mg/L × 250 L = 10,000 mg
1,000 mg = 1 g
∴ 10,000 mg = 10 g
The mass of oxygen needed to be pumped in the aquarium = 10,000 mg = 10 g.
