The spent acid from a nitration process contains 33% h2so4, 36% hno3 and 31% h2o. this is to be strengthened by the addition of concentrated h2so4 containing 95% h2so4 and nitric acid containing 78% hno3. the final strengths of mixed acid solution contains 40% h2so4 43% hno3 and the rest water. calculate : a. the amount of spent acid

b. the amount of concentrate h2so4

c. thr amount of concentrate hno3 to be mixed to produce 1500 kg of tge desired acid

a) 217.15 kg of spent acid

b) 556.14 kg of concentrate H₂SO₄

c) 726.70 kg of concentrate HNO₃

Explanation:

First, let's do a sketch for the problem (image below) and let's choose a basis of 1500 kg of production of the desired acid. Calling A the flow of spent acid, B the amount of concentrated H₂SO₄, C the flow of concentrated HNO₃, and P the flow of the product.

a) The global mass balance is:

input = output

A + B + C = P

A + B + C = 1500

The mass balance for H₂SO₄:

0.33A + 0.95B = 0.4x1500

0.33A + 0.95B = 600

0.95B = 600 - 0.33A

B = (600 - 0.33A)/0.95

The mass balance for HNO₃:

0.36A + 0.78C = 0.43x1500

0.36A + 0.78C = 645

0.78C = 645 - 0.36A

C = (645 - 0.36A)/0.78

Replacing B and C in the global mass balance:

A + (600 - 0.33A)/0.95 + (645 - 0.36A)/0.78 = 1500

0.1911A + 1458.5 = 1500

0.1911A = 41.5

A = 217.15 kg of spent acid

b) B = (600 - 0.33A)/0.95

B = (600 - 0.33x217.15)/0.95

B = 556.14 kg of concentrate H₂SO₄

c) C = (645 - 0.36A)/0.78

C = (645 - 0.36x217.15)/0.78

C = 726.70 kg of concentrate HNO₃

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metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.