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# Tutor profile: Neehal M.

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Neehal M.
Tutor for 3 years and Engineering Consultant
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## Questions

### Subject:Chemistry

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Question:

A solution has a pH = 1. This solution ... ? A) contains no OH─ ions. B) neutralises a hydrochloric acid solution of pH = 1. C) contains a higher concentration of H3O+ ions than OH─ ions. D) contains a higher concentration of OH─ ions than H3O+ ions.

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Neehal M.

C

### Subject:Chemical Engineering

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Question:

It is required to prepare 1250 kg of a solution (M) composed of 12 wt.% ethanol and 88 wt.% water. Two solutions are available, the first stream (A) contains 5 wt.% ethanol, and the second stream (B) contains 25 wt.% ethanol. How much of each solution are mixed to prepare the desired solution?

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Neehal M.

Ethanol Balance: Mass-in = Mass-out A (5/100) + B(25/100) = 0.12M A = (150-0.25B)/0.05 Therefore: A = 3000 - 5B..... (1) Water Balance: Mass-in = Mass-out 0.95A+ 0.75B = 0.88M =0.88( 1250)..... (2) Therefore sub (1) into equation (2): 0.95(3000 - 5B) + 0.75B = 1100 Therefore B = 437.5 kg A = 812.5 kg

### Subject:Physical Science

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Question:

A cat of mass 2.6 kg is sitting on a stationary skateboard (A) of mass 3.5 kg. The cat jumps with a horizontal velocity of 3 m/s to the right and lands on skateboard B (3.5kg) with the same velocity. Skateboard B was also initially at rest prior to the cat landing on it. The surface is frictionless and flat (horizontal). Calculate the velocity of skateboard A after the cat has jumped off it.

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Neehal M.

m1 = mass skateboard A m2 = mass of cat Total linear momentum equation: m1v1 + m2v2 = m1f v1f + m2f v2f (3.5kg)(0) + (2.6kg)(0) = (3.5kg) v1f + (2.6kg)(3 m/s) v1f = -2.23 m/s Therefore the skateboard moves to the left at 2.23 m/s, whilst the cat jumps to the right.

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