1) A carbohydrate sample weighing 0.235 g was found to have a fuel value of 3.84

Question

1) A carbohydrate sample weighing 0.235 g was found to have a fuel value of 3.84 kJ. What is the fuel value of one gram of this carbohydrate, in nutritional Calories? A) 3,910 Cal B) 0.535 Cal C) 16.3 Cal D) 643 Cal E) 3.91 Cal

2) The reaction A + 2B + C ® D + 2E is first order in reactant A, first order in B, and second order in C.
What is the rate law equation for this reaction?
A) rate = k[A][B][C]2 D) rate = [A][2B][C]2
B) rate = [A][B][C]2 E) rate = [A][B]/2[C]
C) rate = [A][B][2C]

3) When a sample of aqueous hydrochloric acid was neutralized with aqueous sodium hydroxide in a
calorimeter, the temperature of 100.0 g of water surrounding the reaction increased from 25.0°C to 31.5°C. If
the specific heat of water is 1.00 cal/(g°C), calculate the quantity of energy in calories involved in this
neutralization reaction.
A) 1000 cal B) 100.0 cal C) 6.50 cal D) 1250 cal E) 650 cal

4)Consider the reversible reaction: A(g) 2B(g)
At equilibrium, the concentration of A is 0.381 M and that of B is 0.154 M. What is the value of the equilibrium
constant, Keq?
A) 0.0622 B) 0.404 C) 1.06 D) 2.47 E) 16.1

5)A buffer solution contains carbonic acid (H2CO3) and sodium bicarbonate (NaHCO3), each at a
concentration of 0.100 M. The relevant equilibrium is shown below. What is the pH of this buffer solution?
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3
(aq) Ka = 4.5 ´ 107
A) 4.50 B) 7.00 C) 6.35 D) 7.65 E) 2.16

Please show Work!!! thank you

Explanation / Answer

1)

Fuel value of food is defined as, “Fuel values are defined as the amount of energy generated by complete combustion of a particular mass of the fuel”

Hence,

           0.235 g of food 3.84 kJ of energy

Then, 1 g of food         say E kJ of energy

By cross multiplication method,

E x 0.235 = 3.48 x 1

E = 3.84/0.235

E = 16.34 kJ

E = 16.34 x 1000 J .............. (1 kJ = 1000 J)

E = 16340 J

E = 16348 x 0.239 cal   ……………..(1 cal = 4.184 J implies 1 J = (4.184)–1 = 0.239 cal)

E = 3907.8 cal

E = 3910 cal   (apprx.)

Option-A

=============================

2) In rate law concentration are raised to the power equal to the their individual order in chemical reaction.

For the reaction, A + 2B + C ---------> D + 2E

Individual order of

A is 1, B is 1, and of C is 2.

Hence the rate law expression is,

Rate = [A]1[B]1[C]2

Option-A

=============================

3)

Enthalpy change i.e. heat absorbed by water (q), temperature change (rise here)(T), specific heat Cp and mass of heat absorber (water here) m in gram are related as,

q = m x Cp x T …………. (1)

For neutralization experiment carried out here,

Mass of water (m) = 100 g

Specific heat of water Cp = 1.00 cal/g oC.

T = 31.5 – 25 = 6.5 oC

Let us put all these values in above eq. (1) and find q

q = 100 g x 1.00 cal/g oC x 6.5 oC

q = 650 cal

Option-E

====================================

4) Given equilibrium reaction is,

A (g) <--------> 2 B (g)

Expression for equilibrium constant (Keq)

Keq = [B]2/[A] ………..(2)

Where [B], [A] are equilibrium concentrations.

Given data:

Equilibrium concentrations are,

[B] = 0.154 M and [A] = 0.381 M

Let us put these values in eq.(1) and find Keq

Keq = (0.154)2/(0.381)

Keq = (0.0237) / (0.381)

Keq = 0.0622

Option-A

=========================================

5) pH of a buffer is given by Henderson- Hasselbalch equation,

pH = pKa + log {[Salt]/[Acid]} ……….(3)

For H2CO3-NaHCO3 (Carbonic acid – Sodium bi-carbonate) buffer,

Both H2CO3 and NaHCO3 are with equal concentration

i.e. [H2CO3] =[NaHCO3] = 1

[H2CO3] =[NaHCO3]

i.e. [NaHCO3] / [H2CO3] = 1

with this values in eq.(3) we get,

pH = pKa + log (1)

pH = pKa + 0 …………..(log 1 = 0)

pH = pKa

But pKa is defined as pKa = - log (Ka)

Hence,

pH = – log(Ka)

For H2CO3 , Ka = 4.5 x 10–7

Put it in above equation

pH = – log (4.5 x 10–7)

pH = 6.35

Option-C

Note: for buffer with equimolar concentrations of acid and salt pH is numerically equal to pKa.

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