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What I answered in a quiz of chemistry and process engineering?

 by  Maryam Hussain

1) How do I make 1 N sodium hydroxide?

Answer: Prepare 1 M (1 molar) solution of sodium hydroxide by dissolving 40 grams of sodium hydroxide in water such that 1 liter of solution is obtained.

2) How many molecules of CO2 are there in 0.05 mol of CO2?

Answer: 3.011 X 1022 molecules

3) Is there proof that A^2 = (A+B) ^2?

Answer: put B = 0, then A^2 = (A+B) ^2

4) Is 0.50 grams the mole of NH4OH?

Answer: 35 grams represents one mole of NH4OH

5) If I dissolve 1kg of sugar in water and I then evaporate the solution will I get 1 kg of sugar?

Answer: Since solubility is due to weak forces & heating, evaporation, for example, breaks these weak force bonds but during evaporation, some of the dissolved solute, sugar, is carried over by rising water vapors. If you recover these evaporated vapors, you can recover sugar.

6) Is mole fraction volume dependent?

Answer: Yes, it is true for ideal gas because both mole & volume fractions are equal.

7) What does the rate of evaporation increase with?

Answer:

a). With liquids having a relatively small surface tension

b). With the increased temperature difference between wall & liquid

c). With an increase in favorable sites on the heat transfer surface

d). With the decreased pressure above the liquid.

8) How much force do you apply?

A huge exhaust fan keeps the inside pressure in a chemistry laboratory room at 8 cm of water vacuum relative to the outside of room. The door of lab has dimensions of 2 m by 1.2 m. What is the average force do you need to open the door? What force did you apply to lift a guy of mass 50 kg by both of your hands? What percentage of force did you apply to lift a guy w.r.t that you applied to open the door of a vacuumed laboratory?

Solution:

The net force on the door is the difference between the forces on the two sides.

Force 1 = (Pressure outside the lab – Pressure inside the lab) x Conversion factor 1 x Conversion factor 2 x Area of door

Force 1 = 8 x  0.01  x  9.8 × 2 × 1.2

F 1 = 1.8816 kN

F 1 = 1881.6 N

Conversion factor 1 : 1 m / 100 cm = 0.01 m / cm

Conversion factor 2 : 1 m H2O =  9.8 kN / m2

The force applied to lift a guy of mass 50 kg by both of your hands is equal to weight of that guy.

Force 2 = Weight = mass of guy x acceleration due to gravity

F 2 = 50 x 9.8

F 2 = 490 N

Percentage of force you need to lift a guy w.r.t that you needed to open the door of a vacuumed laboratory

% F         = 100 – { ( F1 – F2 ) x 100 / F1 }

% F         = 100 – { ( 1881.6 N - 490 N ) x 100 / 1881.6 }

% F         = 26 %   !

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