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Fantom [35]
3 years ago
11

A 30 wt % solution of NaOH is diluted in a mixer to 5 wt%. If the streams entering and leaving the mixer are at 40 oC, find the

heat removed from the mixer on a basis of 100 kg of feed solution.
Chemistry
1 answer:
Strike441 [17]3 years ago
6 0

Explanation:

It is given that 100 kg feed solution  contains 30 wt% solution NaOH.  Hence, feed contains 70 kg of water and 30 kg of NaOH.

Therefore, mixer outlet shows 5 wt% NaOH.

Let us assume that the mixture outlet be F kg  so, by applying mass balance of NaOH we get the value of force as follows.

          F \times 0.05 = 30

                 F = 600 Kg

In that 30 kg is NaOH and 570 kg is water  which also means that initially water present is 70 kg. And, additional water added is 500 kg .

Thus, water feed rate is 500 kg/hr.

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Answer:

pH = 1.85

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The reaction of H₂NNH₂ with HNO₃ is::

H₂NNH₂ + HNO₃ → H₂NNH₃⁺ + NO₃⁻

Moles of H₂NNH₂ and HNO₃ are:

H₂NNH₂: 0.0400L ₓ (0.200mol / L) = 8.00x10⁻³ moles of H₂NNH₂

HNO₃: 0.1000L ₓ (0.100mol / L) = 0.01 moles of HNO₃

As moles of HNO₃ > moles of H₂NNH₂, all H₂NNH₂ will react producing H₂NNH₃⁺, but you will have an excess of HNO₃ (Strong acid).

Moles of HNO₃ in excess are:

0.01 mol - 8.00x10⁻³ moles = 2.00x10⁻³ moles of HNO₃ = moles of H⁺

Total volume is 100.0mL + 40.0mL = 140.0mL = 0.1400L.

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[H⁺] = 2.00x10⁻³ moles / 0.1400L = 0.0143M

As pH = - log [H⁺]

<h3>pH = 1.85 </h3>
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