# Multiple choice question for engineering

## Set 1

1. Which type of graph plot is used to analyze the effect of temperature on the rates of chemical reactions?

a) Dot plot

b) Normal probability plot

c) Arrhenius plot

d) Line weaver-Burk plot

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ln (k) =ln (A) – (E_{a}/R) (1/T)
Where :
k= Rate constant
A= Pre-exponential factor
E_{a}= Activation energy
R= Gas constant
T= Absolute temperature, K
In the right, which represents the y-coordinate axis.

2. From the below representation of growth phase, which is the missing growth phase?

a) Exponential Growth Phase

b) Final Lag Phase

c) Initial Log Phase

d) Final Log phase

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3. If the substrate is given in molar concentration how would you define it’s Molarity and Molality based on the temperature condition?

a) Molarity and Molality both will change on effect of temperature

b) Molarity will change but not Molality

c) Molality will change but not Molarity

d) Normality of the reaction will change

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4. What is the generation time of a bacterial population that increases from 10,000 cells to 10,000,000 cells in four hours of growth?

a) 24 minutes

b) 30 minutes

c) 34 minutes

d) 60 minutes

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^{n}(This equation is an expression of growth by binary fission) Solve for n: Logb = logB + nlog2 n = logb – logB log2 n = logb – logB .301 n = 3.3 logb/B G = t/n

5. On which type of plot the best fit line is represented?

a) Dot plot

b) Line weaver- Burk plot

c) Scatter plot

d) Bland- Altman plot

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6. In the case of Rota meter, which is used for measuring the water flow rate?

a) Only one variable is involved, i.e. independent variable ( the Rota meter reading)

b) Only two variables are involved: one independent variable( the Rota meter reading) and one dependent variable (the water flow rate)

c) Only two variables are involved: one independent variable(the water flow rate) and one dependent variable (the Rota meter reading)

d) Only one variable is involved, i.e. independent variable ( the water flow rate)

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7. The diauxic growth curve mainly found in E.coli in case of two types of nutrients is found in Batch culture is based on which equation?

a) Monod equation

b) Michaelis – Menten equation

c) Schrodinger equation

d) Lyapunov equation

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8. Essentially in an experiment involving bacteria, we measure Optical density of the media where the bacteria grow at different time point from the same flask. If we have 2 or more types of bacteria and want to investigate whether the growth curve of those bacteria are significantly different, what do you think the best analysis can be applied for this purpose?

a) Linear regression analysis or general regression analysis

b) Non- linear regression analysis or general regression analysis

c) Logistic regression analysis

d) Stepwise regression analysis

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9. What is the basic Optical Density (O.D) of the bacterial cells?

a) 680nm

b) 600nm

c) 200nm

d) 280nm

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10. For what MANOVA is used for?

a) Detection of biological samples

b) Detection of the best fit curve

c) Comparison between monovariate samples

d) Comparison between multivariate samples

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## Set 2

1. “Heat transferred to raise or lower the temperature of a material is called Specific heat”.

a) True

b) False

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2. The symbol “Û” refers to?

a) Amount

b) Rate of enthalpy

c) Molar flow rate

d) Specific internal energy

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3. Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole % acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the flowchart shown below.

The process operates at steady state. Calculate the required cooling rate.

Reference states for acetone and nitrogen are-

N_{2} (g, 25°C, 1 atm), Ac (l, 20°C, 5 atm)

a) – 2390kW

b) – 2320kW

c) – 3560kW

d) – 3570kW

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_{2}) Write and simplify the energy balance. Note the following points about the table – (i) Nitrogen has only one state for inlet and outlet. (ii) Acetone has one inlet state but two outlet state (iii) Since the liquid acetone leaving the system is at the reference state, so its specific enthalpy is zero. Calculate all unknown specific enthalpies –

4. Calculate how much energy (kJ) is required to heat a bottle containing 300 mL of liquid water from room temperature (20 °C) to its normal boiling point (100 °C), while still remaining a liquid.

a) 104.4 kJ

b) 100.4 kJ

c) 140.4 kJ

d) 104.5 kJ

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5. Calculate ΔH for a process in which 2.0 mole of NaOH is dissolved in 400 mol H_{2}O at 25C.

(ΔH ̂s at r = 200, 25° C is – 42.26 kJ/ mol)

a) –84.52 kJ

b) -80.42 kJ

c) –64.52 kJ

d) –60. 42 kJ

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6. What term is used for Temperature measured by Thermometer?

a) Wet- bulb temperature

b) Dry- bulb temperature

c) Dew point temperature

d) Normal temperature

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7. Estimate the following properties of humid air at 41°C and 10% relative humidity:

Absolute humidity = 0.0048 kg H2O/ kg DA

Wet-bulb temperature = 19°C

Humid volume = 0.895 M^{3}/ kg DA

Dew point = 3°C

Specific enthalpy = 54.2 – 0.7 = 53.5 kJ/ kg DA

What is the amount of water in 150 m^{3} of air at these conditions?

a) 0.605 kg H_{2}O

b) 0.705 kg H_{2}O

c) 0.805 kg H_{2}O

d) 0.905 kg H_{2}O

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8. What is the enthalpy of 130 g formic acid at 70°C and 1 atm relative to 25°C and 1 atm?

Cp for formic acid in the temperature range of interest is 0.524 cal g-1 °C-1.

a) 4.68 kcal

b) 3.06 kcal

c) 2.06 kcal

d) 2.68 kcal

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ΔH = 3065.4 cal or ΔH = 3.06 kcal

Relative to H=0 at 25°C the enthalpy of formic acid at 70°C is 3.06 kcal.

9. Processes for phase change of vapour to liquid is called?

a) Vaporization

b) Fusion

c) Sublimation

d) Condensation

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10. 50 g benzaldehyde vapour is condensed at 179°C, What is the enthalpy of the liquid relative to the vapour?

Given: The molecular weight of benzaldehyde is 106.12, the normal boiling point is 179.0°C and the standard heat of vaporization is 38.40 kJ gmol-1. For condensation the latent heat is – 38.40 kJ gmol-1.

a) -18.09 kJ

b) -17.09 kJ

c) – 18.06 kJ

d) – 17.06 kJ

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## Set 3

1. Aerobic reactions are not batch operations.

a) True

b) False

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2. In a perfectly mixed reactor _________

a) The output composition is different from input composition

b) The output composition is identical from input composition

c) Both output and input composition are constant

d) Both output and input composition are not constant

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3. Convert v_{max} = 2.5 mmol m^{-3} s^{-1} into mM h^{-1}.

a) 2500

b) 900

c) 25

d) 9

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4. Convert v_{max} = 7 mmol m^{-3} s^{-1} into mM h^{-1}.

a) 25

b) 20

c) 25.20

d) 20.25

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5. In batch reaction time with enzyme deactivation, calculate the first-order deactivation rate constant.

(Given – so = 12 mM; v_{max} = 9 mM h^{-1}; Km = 8.9 mM; sf = 1.2 mM; th = 4.4 h).

a) 0.150 h^{-1}

b) 0.158 h^{-1}

c) 0.155 h^{-1}

d) 0.154 h^{-1}

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6. By taking the parameters of Q5, and the batch reaction time tb.

a) 5.0 h

b) 10.0 h

c) 15.0 h

d) 20.0 h

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7. Zymomonas mobilis is used to convert glucose to ethanol in a batch fermenter under anaerobic conditions. The yield of biomass from substrate is 0.06 g g- 1; YPX is 7.7 g g- 1. The maintenance coefficient is 2.2 g g^{-1} h^{-1}; the specific rate of product formation due to maintenance is 1.1 h^{-1}. The maximum specific growth rate of Z. mobilis is approximately 0.3 h^{-1.5} g bacteria are inoculated into 50 litres of medium containing 12 g l^{-1} glucose. Determine batch culture times required to produce 10 g biomass.

a) 3.1 h

b) 3.3 h

c) 3.5 h

d) 3.7 h

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_{xs}= 0.06 gg

^{-l}; Y

_{px}= 7. 7 gg

^{-l}; μ

_{max}= 0. 3 h

^{-l}; m

_{s}=2.2 gg

^{-l}h

^{-l}; m

_{p}= 1.1 h

^{-l}; x

_{0}=5g/501= 0.1 gl

^{-l};s

_{0}= 12 gl

^{-l}.

If 10 g biomass are produced by reaction, the final amount of biomass present is (10 + 5) g = 15 g. Therefore xf= 15 g/ 50 l = 0.3 g l-1.

8. By taking the pararmeters of Q7 and Determine batch culture times required to achieve 90% substrate conversion.

a) 5.1 h

b) 5.3 h

c) 5.5 h

d) 5.7 h

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9. By taking the parameters question 8 and question 7, determine batch culture times required to produce 100 g ethanol.

a) 3.2 h

b) 3.4 h

c) 3.6 h

d) 3.8 h

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10. The Zymomonas mobilis cells are used for chemostat culture in a 60 m^{3} fermenter. The feed contains 12 g l^{-1} glucose; Ks for the organism is 0.2 g l^{-1}. What flow rate is required for a steady-state substrate concentration of 1.5 g l^{-1}?

a) 15.6 m^{3} h^{-1}

b) 15.8 m^{3} h^{-1}

c) 15.4 m^{3} h^{-1}

d) 15.2 m^{3} h^{-1}

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_{xs}= 0.06 g g

^{-1}; Y

_{px}= 7.7 g g

^{-1};μ

_{max}= 0. 3 h

^{-1}; K

_{s}= 0.2 g

^{-1}; m

_{s}= 2.2 g g

^{-1}h-l; s

_{i}= 12 g 1

^{-1}; V=60 m

^{3}. qp = 3.4 h

^{-1}, YPS = Ypx Yxs = 0.46 g g

^{-1}.

s = 1.5 g l^{-1}

11. By taking Q10 and at the flow rate of Q10, what is the cell density?

a) 0.42 g l^{-1}

b) 0.44 g l^{-1}

c) 0.46 g l^{-1}

d) 0.48 g l^{-1}

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12. By taking Q10 and Q11 into account and at the flow rate of Q10, what concentration of ethanol is produced?

a) 5.1 g l^{-1}

b) 5.3 g l^{-1}

c) 5.5 g l^{-1}

d) 5.7 g l^{-1}

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13. Which of the following type is of the perfusion culture?

a) Batch

b) Conc. Batch

c) Continuous

d) Conc. Fed-Batch

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14. Immobilised lactase is used to hydrolyse lactose in dairy waste to glucose and galactose. Enzyme is immobilised in resin particles and packed into a 0.5 m^{3} column. The total effectiveness factor for the system is close to unity; Km for the immobilised enzyme is 1.32 kg m^{-3}; Vmax is 45 kg m^{-3} h^{-1}. The lactose concentration in the feed stream is 9.5 kg m^{-3}; a substrate conversion of 98% is required. The column is operated with plug flow for a total of 310 d per year. At what flow rate should the reactor be operated?

a) 1.56 m^{3} hh^{-1}

b) 1.58 m^{3} hh^{-1}

c) 1.50 m^{3} h-1

d) 1.54 m^{3} h-1

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^{-3}

15.By taking the parameters of Q14 into consideration, estimate how many tonnes of glucose are produced per year?

a) 56.3 tonnes yr^{-1}

b) 56.6 tonnes yr^{-1}

c) 56.7 tonnes yr^{-1}

d) 56.5 tonnes yr^{-1}

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F(si – sf)= 1.56 m^{3} h^{-1} (9.5 – 0.19) kg m^{-3}= 14.5 kg h^{-1}

Converting this to an annual rate based on 310 d per year and a molecular weight for lactose of 342:

The enzyme reaction is:
lactose + H_{2}0 → glucose + galactose.
Therefore, from reaction stoichiometry, 315 kg mol glucose are produced per year. The molecular weight of glucose is 180; therefore:

## Set 4

1. Which is the correct order for the equation of mass balance of the system?

{A} – {B} + {C} – {D} = {Mass accumulated within system}

a) A. Mass consumed within system, B. Mass in through system boundaries, C. Mass out through system boundaries, D. Mass generated within system

b) A. Mass generated within system, B. Mass in through system boundaries, C. Mass out through system boundaries, D. Mass consumed within system

c) A. Mass in through system boundaries, B. Mass out through system boundaries, C. Mass generated within system, D. Mass consumed within system

d) A. Mass consumed within system, B. Mass generated within system, C. Mass out through system boundaries, D. Mass in through system boundaries

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2. If the Reactant iron is combined with Reactant Sulfur to form the product Iron sulfide, then what will be the atomic mass of the product?

a) 88

b) 44

c) 22

d) 80

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3. If the reactant Magnesium reacts with Reactant Hydrochloric acid to form the product Magnesium chloride + Hydrogen, then what will be the atomic mass of both the product?

a) 90

b) 97

c) 80

d) 87

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4. A continuous process is set up for treatment of wastewater. Each day, 10^{3} kg cellulose and 10^{5} kg bacteria enter in the feed stream, while 102 kg cellulose and 1.5 x 10^{2} kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 102 kg d^{-1}. The rate of bacterial growth is 4x 10^{2} kg d^{-l}; the rate of cell death by lysis is 6 x 10^{2} kg d ^{-1}. Write balances for cellulose and bacteria in the system.

a) 1× 103 kg, 9× 10^{3} kg

b) 1× 102 kg, 9.965 × 10^{4} kg

c) 1× 103 kg, 9.964 × 10^{4} kg

d) 1× 102 kg, 9× 10^{3} kg

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^{3}– 10

^{2}+ 0 – 8 x 10

^{2}) = accumulation

Therefore, 1× 10^{2} kg cellulose accumulates in the system each day.

Performing the same balance for bacteria:

(10^{5} – 1.5 x 10^{2} + 4x 10^{2} – 6 x 10^{2}) = accumulation

Therefore, 9.965 × 10^{4} kg bacterial cells accumulate in the system each day.

5. Which of the following system follows the differential equation?

a) Semi- Batch process

b) Batch process

c) Fed- Batch process

d) Continuous process

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6. 2SO_{2}+O_{2}→ 2SO_{3}. What is the stoichiometric ratio of SO_{2} to SO_{3}?

a) 3

b) 1

c) 2

d) 0

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_{2}to SO

_{3}= (2 mole of SO

_{2}3 produced) = 1.

7. Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt% methanol and the second contains 70.0 wt% methanol. If 200 kg of the first mixture is combined with 150 kg of the second, what are the mass and composition of the product (Total balance, Methanol balance, Physical constraint) ?

a) 350 kg, 0.529, 0.471

b) 320 kg, 0.529, 0.521

c) 360 kg, 0.620, 0.471

d) 350 kg, 0.690, 0.760

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Total Balance: m_{1} + m_{2} = m_{3} Methanol-Balance: m_{1}x_{M1} + m_{2}x_{M2} = m_{3}x_{M3} Water-Balance: m_{1}xw1 + m2xw2 = m3xw3

(choose only 2 equations since one of them is no longer independent)

Physical Constraint (applied to mixture 3):

x_{M3} + x_{W3} = 1.00

Always start with the equation with the least number of unknowns if possible and minimize solving equations simultaneously.

Total Balance (m_{3})
↓
Methanol Balance (x_{M3})
↓
Physical Constraint (x_{W3})
Total balance:
m_{3} = (200 kg) + (150 kg)
m_{3} = 350 kg

CH3OH balance:
(200 kg)(0.40) + (150 kg)(0.70) = (350 g)x_{M3} x_{M3} = 0.529

Physical constraint:
x_{W3} = 1.00 – x_{M3} = 1 – 0.529
x_{W3} = 0.471.

8. A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m^{3}. Calculate the concentration of salt in this solution as a (a) weight fraction, (b) weight/volume fraction, (c) mole fraction, (d) molal concentration.

a) 15.7%, 23.1%, 0.058, 3.88 moles in m^{3}

b) 14.5%, 22.4%, 0.059, 2.88 moles in m^{3}

c) 16.7%, 22.1%, 0.058, 3.77 moles in m^{3}

d) 14.7%, 23.1%, 0.059, 2.77 moles in m^{3}

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20/(100+20) = 0.167; %weight/weight = 16.7%

(ii) Weight/volume:

A density of 1323kg/m^{3} means that lm3 of solution weighs 1323kg, but 1323kg of salt solution contains

( 20×1323kg of salt) / (100+20) = 220.5 kg salt/m3

1 m^{3} solution contains 220.5 kg salt.
Weight/volume fraction = 220.5 / 1000 = 0.2205
And so weight / volume = 22.1%

(iii) Moles of water = 100 / 18 = 5.56 Moles of salt = 20 / 58.5 = 0.34 Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058

(iv) The molar concentration (M) is 220.5/58.5 = 3.77 moles in m^{3}.

9. Calculate the degree of reduction for ethanol, where the degree of reduction of C=4, H=1,

O= -2.

a) ϒ = 9

b) ϒ = 6

c) ϒ = 3

d) ϒ = 2

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Explanation: Ethanol (C2H5OH) = 2(4) + 6(1) + 1(-2) = 12, ϒ = 12/2 = 6

As, The number of equivalents of available electrons per gram atom C is the measure of degree of reduction.

10. Propane (C_{3}H_{8}) burns in this reaction:

C_{3}H_{8} + 5O_{2} = 4H_{2}O + 3CO_{2}

If 200 g of propane is burned, how many g of H_{2}O is produced?

a) 327.27 g

b) 345.5 g

c) 323.2 g

d) 232.3 g

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Since you cannot calculate from grams of reactant to grams of products you must convert from grams of C_{3}H_{8} to moles of C_{3}H_{8} then from moles of C_{3}H_{8} to moles of H_{2}O. Then convert from moles of H_{2}O to grams of H_{2}O.

Step 1: 200 g C_{3}H_{8} is equal to 4.54 mol C_{3}H_{8}.
Step 2: Since there is a ratio of 4:1 H_{2}O to C_{3}H_{8}, for every 4.54 mol C_{3}H_{8} there are 18.18 mol H_{2}O.
Step 3: Convert 18.18 mol H_{2}O to g H_{2}O. 18.18 mol H_{2}O is equal to 327.27 g H_{2}O.

## Set 5

1. If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:

(i) the mean molecular weight of air,

(ii) the mole fraction of oxygen,

(iii) the concentration of oxygen in mole/m^{3} and kg/m^{3} if the total pressure is 1.5 atmospheres and the temperature is 25 degreeC.

a) 29.8, 0.21, 0.010 mole/m^{3}

b) 28.8, 0.21, 0.013 mole/m^{3}

c) 27.6, 0.30, 0.010 mole/m^{3}

d) 29.8, 0.30, 0.013 mole/m^{3}

### View Answer

_{2}and 23/32 moles of O

_{2},

Total number of moles = 2.75 + 0.72 = 3.47 moles.
So mean molecular weight of air = 100 / 3.47 = 28.8
Mean molecular weight of air = 28.8
(ii) The mole fraction of oxygen = 0.72 / (2.75 + 0.72) = 0.72 / 3.47 = 0.21
Mole fraction of oxygen = 0.21
(iii) In the gas equation, where n is the number of moles present: the value of R is 0.08206 m3 atm/mole K and at a temperature of 25 ⁰C = 25 + 273 = 298 K, and where V= 1 m^{3} pV = nRT
and so, 1.5 x 1 = n x 0.08206 x 298 n = 0.061 mole/m^{3} weight of air = n x mean molecular weight = 0.061 x 28.8 = 1.76 kg / m^{3} and of this 23% is oxygen, so weight of oxygen = 0.23 x 1.76 = 0.4 kg in 1 m^{3} Concentration of oxygen = 0.4kg/m^{3} or 0.4 / 32 = 0.013 mole / m^{3}

When a gas is dissolved in a liquid, the mole fraction of the gas in the liquid can be determined by first calculating the number of moles of gas using the gas laws, treating the volume as the volume of the liquid, and then calculating the number of moles of liquid directly.

2. In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0⁰C and atmospheric pressure. Calculate (i) the mass fraction and (ii) the mole fraction of the CO_{2} in the drink, ignoring all components other than CO_{2} and water.

Basis 1 m^{3} of water = 1000 kg

Volume of carbon dioxide added = 3 m^{3}

From the gas equation, pV = nRT

1 x 3 = n x 0.08206 x 273 n = 0.134 mole.

Molecular weight of carbon dioxide = 44

And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg

a) 6.9 × 10-3, 3.41 × 10^{-3}

b) 8.9 × 10-3, 2.41 × 10^{-3}

c) 5.9 × 10-3, 2.41 × 10^{-3}

d) 9.9 × 10-3, 3.41 × 10^{-3}

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^{-3}(ii) Mole fraction of carbon dioxide in drink = 0.134 / (1000/18 + 0.134) = 2.41 × 10

^{-3}.

3. Each year 80,000 people move into a city, 70,000 people move out, 18,000 are born, and 11,000 die. Write a balance on the population of the city.

a) 17,000 P/yr

b) 15,000 P/yr

c) 20,000 P/yr

d) 22,000 P/yr

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4. According to the Balances on steady-state processes, the accumulation is equal to?

a) 1

b) 0

c) 100

d) 2

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5. A process unit involves 3 chemical components. How many mass balances can be written?

a) 3

b) 5

c) 6

d) 4

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6. The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor.

C_{2}H_{4} + HBr = C_{2}H_{5}Br

The product stream is analyzed and found to contain 51.7 mole% C_{2}H_{5}Br and 17.3% HBr. The feed to the reactor contains only ethylene and hydrogen bromide.

Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is 165 mol/s, what is the extent of reaction?

a) 56.2 mol/s

b) 45.6 mol/s

c) 55.6 mol/s

d) 44.6 mol/s

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7. What is the fractional and percent excess of H_{2} from the following equation:

H_{2} + Br_{2} = 2HBr

H_{2}= 25 mol/hr

Br_{2}= 20 mol/hr

a) 0.45, 45%

b) 0.25, 25%

c) 0.75, 75%

d) 0.65, 65%

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_{2}= (25-20)/ 20 = 0.25 Percent excess = 25%.

8. When pure carbon is burned in air, some of it oxidizes into CO_{2} and some to CO. The molar ratio of N_{2} to O_{2} is 7.18 and the ratio of CO to CO_{2} is 2.0 in the product gas. What is the percent excess air used? The exit gases contain only N_{2}, O_{2}, CO and CO_{2}.

a) 50 %

b) 60 %

c) 40 %

d) 20 %

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_{2}= CO

_{2}R2 = C + ½ O2 = CO As the product (output) gas composition is specified more clearly, we may use it as starting point. Take N

_{2}in the product as nN2 = Take N2 in the product as, then O

_{2}= 1 mol N

_{2}as an inert gas: input = output = 7.18 mol O2 input with air = 7.18 (21/79) = 1.91 mol Total O

_{2}consumption = 1.91-1 = 0.91 mol If R1 uses n1 mol O

_{2}, generating n1 mol CO

_{2}, R

_{2}uses 0.91-n1 mol O

_{2}, generating 2 (0.91-n1) mol CO. Since CO/CO2 in the product gas = 2, 2 (0.91-n1)/n1 = 2 n1 = 0.91/2 = 0.455 mol Therefore, total moles of C input = 3 n1 = 1.365 mol O

_{2}required for complete combustion of 1.365 mol C = 1.365 mol (for R1 only) O

_{2}excess % = (1.91 -1365)/1.365 x 100%= 39.9% =40% .

9. Convert 800 mmHg into bars.

a) 1.064 bars

b) 1.066 bars

c) 1.054 bars

d) 1.055 bars

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^{5}Pa) = 1.066 bar

10. What do you mean by material balance equation?

a) Original-Solids-in-place (OSIP)

b) Original-Gas-in-place (OGIP)

c) Original-Air-in-place (OAIP)

d) Original-Liquid-in-place (OLIP)