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JAMB UTME - Chemistry - 2017

Vraag 1 Verslag

The shape of the S-orbital is

elliptical

spiral

circular

spherical

Antwoorddetails

The shape of the S-orbital is spherical. This means that it has a symmetrical, 3-dimensional shape that resembles a ball or a globe. It is not flat or elongated, but rather is evenly rounded in all directions. This is the most basic type of orbital and is used to describe the electrons in the outermost energy level of an atom.

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Atomic Structure And Bonding

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Vraag 15 Verslag

The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?

0.1

0.5

2.0

4.0

Antwoorddetails

The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.

R ∝ $\frac{1}{\sqrt{M m}}$ or R ∝ $\frac{1}{\sqrt{D}}$

Dx = 0.5gdm-3, Dy = 2gdm-3

R= $\frac{K}{\sqrt{D}}$

R $\sqrt{D}$ = k

R1 $\sqrt{D_{1}}$ = R1 $\sqrt{D_{2}}$

Rx $\sqrt{D_{x}}$ = Ry $\sqrt{D_{y}}$

$\frac{R_{x}}{R_{y}}$ = $\frac{\sqrt{D_{y}}}{\sqrt{D_{x}}}$

= $\frac{\sqrt{2}}{\sqrt{0.5}}$

= 2.0

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Kinetic Theory Of Matter And Gas Laws

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Vraag 16 Verslag

Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]

0.300g

0.250g

0.2242g

0.448g

Antwoorddetails

M =

\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}

= $\frac{M m I T}{96500 n}$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $\frac{64 \times 1350}{96500 \times 2}$

Mass = 0.448g

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Electrolysis

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Vraag 19 Verslag

For a general equation of the nature xP + yQ ? mR + nS, the expression for the equilibrium constant is

k [P]x [Q]y

\frac{[P]^{x} [Q]^{y}}{[R]^{m} [S]^{n}}

\frac{[R]^{m} [S]^{n}}{[P]^{x} [Q]^{y}}

\frac{m [R] n [S]}{x [P] y [Q]}

Antwoorddetails

Expression for equilibrium constant

k = $\frac{concentration of product}{concentration of reactant}$

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Chemical Equilibria

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Vraag 26 Verslag

Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]

2.536

1.623

4.736

0.394

Antwoorddetails

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = $\frac{R V}{R T}$

n = $\frac{6 \times 105}{0.082 \times 303}$

n= 2.536mol

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Kinetic Theory Of Matter And Gas Laws

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Vraag 34 Verslag

X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be

negative

zero

positive

indeterminate

Antwoorddetails

$\begin{array}{cc} Enthalpy Change [ΔH] & Entropy Change [ΔS] & Gibbs free Energy[ΔG] \\ Positive & Positive & depends on T, may be + or - \\ Negative & Positive & always negative \\ Negative & Negative & depends on T, may be + or - \\ Positive & Negative & always positive \end{array}$

?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.

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Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

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Vraag 37 Verslag

CH2S(g) + O2(g) → 2Cn + SO2(g)

What is the change in the oxidation number of copper in the reaction?

0 to + 2

0 to + 1

C + 1 to 0

+ 2 to + 1

Antwoorddetails

In the reactant;

Cu2S

2 Cu - 2(1) = 0

2 Cu = 2

Cu = $\frac{2}{2}$

Cu = +1

In the product, Cu

Cu = O

The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively

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Oxidation And Reduction

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Vraag 39 Verslag

A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

1.6

\times

10-4

6.3

\times

10-11

6.3

\times

10-4

1.6

\times

10-11

Antwoorddetails

The pH scale measures the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the concentration of the hydronium ion (H3O+), which is given by the expression: pH = -log[H3O+] To calculate the concentration of the hydroxide ion (OH-) in the juice, we need to use the equation for the ion product of water: Kw = [H3O+][OH-] = 1.0 × 10-14 where Kw is the ion product constant of water. At 25°C, Kw has a value of 1.0 × 10-14. To solve for [OH-], we need to rearrange the equation: [OH-] = Kw / [H3O+] Now we can substitute the pH of the orange juice to calculate the concentration of hydronium ion [H3O+]. pH = -log[H3O+] 3.80 = -log[H3O+] [H3O+] = 10^(-3.80) [H3O+] = 1.58 × 10^(-4) mol/L Substituting this value into the ion product equation: [OH-] = Kw / [H3O+] [OH-] = 1.0 × 10^-14 / (1.58 × 10^-4) [OH-] = 6.33 × 10^-11 mol/L Therefore, the concentration of the hydroxide ion in the orange juice is 6.33 × 10^-11 mol/L.

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Acids, Bases And Salts

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