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Alina [70]
3 years ago
8

Explain the purpose of the heat exchanger in the nuclear power plant.

Chemistry
2 answers:
Ymorist [56]3 years ago
5 0

Answer:

A heat exchanger is device which is designed to transfer heat from one matter to another. In nuclear power plants, these heat exchangers are used to pass heat from primary system to the secondary system. In the process they produce steam from water, they are also called steam generators or pressurized water reactors.

My name is Ann [436]3 years ago
3 0

Answer:

yeah someone explain

Explanation:

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The interdependence of agriculture and industry helps the development of both the sectors. The most important aspect of this inter dependence is that the products of one serve as important inputs for the other. Growth of one sector, thus means ample supply of inputs for the other.

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How many moles are in 30g of C12H22O11?
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342.29648 grams

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The simplest type of stream cipher, one in which one letter or character is exchanged for another, is known as what?
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What would happen to the temreture if there was a sudden icrease in the amount of greenhouse gasses
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The higher concentrations of greenhouse gases—and carbon dioxide in particular—is causing extra heat to be trapped and global temperatures to rise.

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2 years ago
An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid,
natima [27]

Answer:

The answer is "Option B".

Explanation:

\to CH_3COOH + NaOH \longleftrightarrow  CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\  NaOH = (5 \ E-3\  L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\

\to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\  L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\

\to C \ CH_3COONa =  \frac{(0.01\  mol + 5 \ E-4\  mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\

\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\  E-6 - 1.75\  E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to  [H_3O^+] = 1.5835\  E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to  pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7

5 0
3 years ago
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