If more powdered Kool-Aid is added to the same amount of water, what happens to the solution (the answer I clicked I don’t know
if correct or not)
1 answer:
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Answer:
Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)
Explanation:
The given parameters are;
The metals provided to melt the ice and their temperature includes;
One kg (1000 g) of iron;
Specific heat capacity = 0.444 J/(g·°C)
Temperature = 100°C
1 kg (1000 g) of lead
Specific heat capacity = 0.160 J/(g·°C)
Temperature = 100°C
Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;
For iron, we have;
ΔQ = Mass × Specific heat capacity × Temperature change
ΔQ = Heat obtained from the iron by the ice
ΔQ = 0.444 m × 1000 × (100 - 0) = 44400 J
Heat absorbed by the ice for melting, H = Heat obtained from the iron
∴ Heat absorbed by the ice for melting, H = Mass of ice × Latent heat of ice
H = Mass of ice × 334 J/g = 44400 J
Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g
Mass of ice melted by the iron ≈ 132.9 g
For lead, we have;
ΔQ = Mass × Specific heat capacity × Temperature change
ΔQ = Heat obtained from the iron by the ice
ΔQ = 0.160 m × 1000 × (100 - 0) = 16000 J
Heat absorbed by the ice for melting, H = Heat obtained from the iron
∴ Heat absorbed by the ice for melting, H = Mass of ice × Latent heat of ice
H = Mass of ice × 334 J/g = 16000 J
Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g
Mass of ice melted by the lead ≈ 47.9 g
Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.
Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;
From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be
and the total pressure at the final equilibrium be
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;
From the foregoing; since the volume is decreased to one- half to initial Volume; then ,
also;
Thus ;
Dividing both sides by
From ;
Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.