An investigation is conducted into how the mass of magnesium metal reacting with hydrochloric acid affects the amount of hydroge
n gas produced.
Masses of 0.10g, 0.20g, 0.30g and 0.40g of powdered Mg metal are reacted with hydrochloric acid(HCl). The conical flask containing the reaction mixture of Mg and HCl is connected to another conical flask with a hose. The hydrogen gas produced by the reaction is collected in this conical flask. The greater the amount of hydrogen gas produced by the reaction the greater the pressure of the gas in the flask.
Answer:
A) what is the independent variable:
B) what is the dependent variable:
C) write a hypothesis for this investigation:
D) give 2 variables that should have been controlled for this investigation:
Masses of 0.10g, 0.20g, 0.30g and 0.40g of powdered Mg metal are reacted with hydrochloric acid(HCl). The conical flask containing the reaction mixture of Mg and HCl is connected to another conical flask with a hose. The hydrogen gas produced by the reaction is collected in this conical flask. The greater the amount of hydrogen gas produced by the reaction the greater the pressure of the gas in the flask.
Answer:
A) what is the independent variable:
B) what is the dependent variable:
C) write a hypothesis for this investigation:
D) give 2 variables that should have been controlled for this investigation:
1 answer:
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Answer:
the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.
Explanation:
Let state the given parameters;
Let A= solvent (hexane)
B= solid(inert soiid)
C= solvent(oil)
= mass of solvent + mass of oil (i.e A+C)
<u>Feed Phase:</u>
Total feed (i.e slurry of flakes soybeans)= 100kg
B= mass of solid =75 kg
F= mass of solvent + mass of oil (i.e A+C)
= 25kg
Mass ratio of oil to solution =
mass of oil (C) =25 × 0.1 wt = 2.5kg
mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg
therefore =
= 0.1
mass ratio of solid to solution =
=3
<u>Solvent Phase:</u>
C= Mass of oil= 0(kg)
A= Mass of hexane = 100kg
mass of solutions = A+C = 0+100kg
solvent= 100kg
<u>Underflow:</u>
underflow = L₁ = (unknown) ???
L₁ = E₁ + B
the value of N for the outlet and underflow is 1.5 kg
i.e N₁ =
solution in underflow E₁ = Mass (A+C)
<u>Overflow:</u>
Overflow = V₁ = (unknown) ???
solution in overflow V₁ = Mass (A+C)
This is because, B = 0 in overflow
Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).
solid in feed = solid in underflow = 75
75= E₁ × N₁
75 = E₁ × 1.5
E₁ = 50kg
Underflow L₁ = E₁ × B
= 50 + 75
=125kg
The Overall Balance: Feed + Solvent = underflow + overflow
100 + 100 = 125 + V₁
V₁ = 75kg