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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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Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,
vfiekz [6]

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Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

8 0
3 years ago
You have 100 mL of a 12M solution of HCl, and you need to dilute it to 1.5M for an experiment. How many liters will your new sol
sashaice [31]

Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

M after dilution = 1.5 M, V after dilution = ??? mL.

∵ (MV)before dilution of HCl = (MV)after dilution of HCl

∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)

<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>

8 0
3 years ago
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