Correct Answer: D. 0.8Ω

Explanation

R = V/I=12/15 = 0.8Ω

Correct Answer: B. repetition

Correct Answer: C) triethylamine

Explanation

To determine which one is the tertiary amine, you need to count the number of ethyl groups attached to the nitrogen atom. If there are three ethyl groups, then it is a tertiary amine. If there are two ethyl groups, then it is a secondary amine. If there is only one ethyl group, then it is a primary amine.

Option A, ethylamine, has only one ethyl group attached to the nitrogen atom, which makes it a primary amine. Option B, diethylamine, has two ethyl groups attached to the nitrogen atom, which makes it a secondary amine. Option C, triethylamine, has three ethyl groups attached to the nitrogen atom, which makes it a tertiary amine and the correct answer. Option D, tetraethylamine, has four ethyl groups attached to the nitrogen atom, which makes it a quaternary ammonium salt, not a tertiary amine.

In summary, the correct answer to this question is Option C: triethylamine, as it is the only option that has three ethyl groups attached to the nitrogen atom, making it a tertiary amine

Correct Answer: D. existing administration farvoured the system

Correct Answer: C. kilowatt-hour

Correct Answer: B. serve Allah without associates

Correct Answer: C. believe in the only Son of God

Explanation

(a)(i) Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons. It expressed in atomic mass unit (a.m.u.). It is also a measure of binding energy.

Mass defect = total mass of nucleus

– mass of nucleons.

(ii) Binding energy is the work done or energy needed to separate the nucleons in a nucleus. It is measured in eV or Joules (J).

(b)(i) The effect of decreasing the speed of a photoelectron on: (\(\alpha\)) wavelength --> wavelength increases. (\(\beta\)) energy ---> energy decreases.

(ii) \(\lambda = \frac{h}{mv}\)

= \(\frac{6.6 \times 10^{-34}}{4.4 \times 10^{-2} \times 10^{-16}}\)m

(c) Pairs of isotopes in the figure are;

\(^{208}_{82}Pb\) and \(^{212}_{82}Pb\) or \(^{212}_{84}Po\) and \(^{216}_{84}Po\) or

\(^{224}_{88}Ra\) and \(^{288}_{88}Ra\) or \(^{288}_{90}Th\) and \(^{232}_{90}Th\)

(ii) The series started with isotopes \(^{232}_{90}Th\)

(iii) Nuclear equation for;

\(\alpha\)Alpha decay

\(^{232}_{90}Th\) \(\to\) \(^{228}_{88}Ra\) + \(^{4}_{2}He\) + Energy;

\(^{228}_{90}Th\) \(\to\) \(^{22}_{88}Ra\) + \(^{4}_{2}He\) + Energy;

\(^{224}_{88}Ra\) \(\to\) \(^{220}_{86}Rn\) + \(^{4}_{2}He\) + Energy;

\(^{212}_{84}Po\) \(\to\) \(^{208}_{82}Pb\) + \(^{4}_{2}He\) + Energy;

\(^{216}_{84}Po\) \(\to\) \(^{212}_{82}Pb\) + \(^{4}_{2}He\) + Energy;

\(^{212}_{84}Rn\) \(\to\) \(^{208}_{82}Po\) + \(^{4}_{2}He\) + Energy;

\(^{220}_{84}Rn\) \(\to\) \(^{216}_{84}Po\) + \(^{4}_{2}He\) + Energy;

\(^{216}_{84}Rn\) \(\to\) \(^{212}_{82}Pb\) + \(^{4}_{2}He\) + Energy;

(\(\beta\)) Beta decay

\(^{224}_{88}Ra\) \(\to\) \(^{224}_{89}Ac\) + \(^{o}_{-1}e\) + Energy;

\(^{228}_{88}Ac\) \(\to\) \(^{228}_{90}Ac\) + \(^{o}_{-1}e\) + Energy;

\(^{212}_{82}Pb\) \(\to\) \(^{212}_{83}Bi\) + \(^{o}_{-1}e\) + Energy;

\(^{212}_{83}Ra\) \(\to\) \(^{212}_{84}Ac\) + \(^{o}_{-1}e\) + Energy;

Correct Answer: B. F

Explanation

Hours of work per day = \(\frac{Income}{wage rate}\)

(a)(i) \(\frac{$10}{$10}\) = 1 hour

(ii) \(\frac{$60}{$20}\) = 3 hours

(iii) \(\frac{$10}{$10}\) = 1 hour

income = per day = wage rate x hours worked

(b) (i) $40 x 6 hours = $ 240

$50 x 4 hours = $200

$60 x 2 hours = $120

(c) (i) $40 per hour

(ii) Backward Bending Supply Curve

(iii) As the wage rate increases from $10 to $40 per hour, the worker increases the number of hours worked from 1 hour to 6 hours a day. The supply curve over this stretch is quite normal as the worker substitues worl for leisure in response to higher wage rate offered. When the highest income is attained at the wage of the $40 the supply curve bends backwards towards the price axis as the labour market experiences the income effect. After that point, the worker puts in fewer hours and trades off extra pay for more leisure.