Correct Answer: B. James

Explanation

9, Aderopo Close,

Oke-Okun,

lbadan

16' July, 2015

Dear James,

In response to your previous message on why you think luck has been what makes my school excellent in academic work, I'd like to highlight the reasons in which my school remains unshakeable when talking about academic excellence.

Would you believe it if I told you that all the teachers in my school have at least, a master's degree? Yes! The school authority saw to it that all teaching staff had at least one master's degree in the disciplines they teach. This has in numerous ways contributed to my school's academic excellence. During classes, it's like you're listening to the modern version of Albert Einstein. Lectures are usually interactive and trust me; you wouldn't want to leave the class during lecture hours. It's always like you're glued to your seat. You get to ask questions and the assignments aren't too much. Understanding is inevitable with the teachers that teach in my school.

Another factor that has contributed to the academic excellence in my school is the use of modern educational facilities. Going to school is like paradise on earth. There are three libraries, twelve computer rooms, four laboratories; in fact, our school just acquired a fish pond capable of housing diverse species of fish.

In every classroom, there are two air conditioners, a digital board; every desk in the room has a tablet (which means we do not use notebooks). Tell me, how won't we excel academically with these things around? Excellence is just inevitable.

Here is the part I love most, our school curriculum is arranged in such a way that students find learning interesting. Subjects are made understandable. There is no day I'd go to school without learning something new. The subjects being taught are broken into absorbable bits. My school remains the best in terms of academic excellence and it'll keep being the best.

I do hope you now understand why my school's success in academic school work is not as a result of luck, I hope to hear from you soon and do send my regards to Juliana.

Yours sincerely,

Ben.

Correct Answer: B. confrontation

Correct Answer: B. 264 ms-1

Explanation

Here's how to determine the velocity of the transverse wave:

1. Understand the fundamental frequency: For a stretched string fixed at both ends, the fundamental frequency (first harmonic) occurs when the string vibrates in a single loop. This means the length of the string (L) is equal to half a wavelength (λ/2).

2. Calculate the wavelength (λ):

Given L = 0.6 m

For the fundamental frequency, L = λ/2

So, λ = 2 * L

λ = 2 * 0.6 m

λ = 1.2 m

3. Use the wave speed formula:

The velocity (v) of a wave is given by the product of its frequency (f) and wavelength (λ):

v = f * λ

4. Substitute the given values:

Given f = 220 Hz

Calculated λ = 1.2 m

v = 220 Hz * 1.2 m

v = 264 m/s

Therefore, the velocity of the transverse wave in the string is 264 ms-1.

The correct answer is B. 264 ms-1.

Correct Answer: B. have a minimum of 100 shareholders

Correct Answer: B. waste a lot of productive energy roaming about

Correct Answer: C. ensures that only dry steam gets into the calorimeter

Correct Answer: B. 45^o

Explanation

In In the diagram given, \(\alpha\) = 22^o (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)

< PSQ + 133^o + \(\beta\) = 180^o (sum of angles of a \(\Delta\))

< PSQ + 133^o + 22^o = 180^o

< PSQ = 180^o - (133 + 22)^o

45^o

Explanation

(a) (i) X - oxygen

Y - sulphur

(ii) I. - S(s) + O\(_{2(g)}\) \(\to\) SO\(_{2(g)}\) (1)

II. - 2SO\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2SO\(_{3(g)}\) (2)

III. - SO\(_{3(g)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\) H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) (1)

IV. - H\(_2\)S\(_2\)O\(_7\)\(_{(l)}\) + H\(_2\)O\(_{(l)}\) 2H\(_2\)SO\(_{4(aq)}\) (2)

(iii) Vanadium (V) oxide (1)

(iv) It is because the reaction is exothermic

(v) - In bleaching wood pulp, wool silk, hay

- as germicide and fumigant/ disinfectant

- as refrigerant / refrigerating gas/coolant

- as preservative for grains and fruits

- manufacture of H\(_2\)SO\(_4\)

- as a reducing agent in metallurgy

- for dechlorination

(b) From the equation,

2 mol of H\(_2\) reacts with 1 mol of O\(_2\)

hence 40 cm\(^3\) of H\(_2\) would react with 20 cm\(^3\) of O\(_2\)

Volume of unused oxygen = (30 – 20) cm\(^3\)

= 10 cm\(^3\)

(c) (i) CaCO\(_3\) \(\to\) CaO + CO\(_2\)

(ii) CaCO\(_3\) = 40 + 12 + (16 x 3) = 100 g

CaO = 40 + 16 = 56 g

100 g of CaCO\(_3\) \(\to\)&

Correct Answer: D. Industrial goods