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o-na [289]
3 years ago
15

The amount of energy required to heat water for a 10-minute shower (50 gallons) is 2.2125 kJ. How many calories is this?

Chemistry
1 answer:
Zigmanuir [339]3 years ago
6 0
From the problem statement, this is a conversion problem. We are asked to convert from units of kilojoules to units of calories. To do this, we need a conversion factor which would relate the different units involved. We either multiply or divide this certain value to the original measurement depending on what is asked. From literature, we will find that 1 kilojoule is equal to 239 calories. We do as follows:
<span>
2.2125 kJ ( 239 calories / 1 kJ ) = 528.79 calories
</span><span>
</span>
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meriva

Answer: An atom can be considered unstable in one of two ways. If it picks up or loses an electron, it becomes electrically charged and highly reactive. Such electrically charged atoms are known as ions. Instability can also occur in the nucleus when the number of protons and neutrons is unbalanced.

Explanation:

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6. Explain the impact on fresh water of:
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Fresh water pollutants  are substances which pollute fresh water and  industrial waste, is most harmful fresh water pollutant to man and aquatic organisms.

<h3>What are pollutants?</h3>

Pollutants are substances which cause harm when they are present in the environment.

Pollutants include chemicals such as petroleum and material such as sewage.

The presence of pollutants in freshwater results in water pollution and make the water unfit for drinking purposes and also harms aquatic life in freshwaters.

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1 year ago
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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -&gt; 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
3 years ago
A proton is moving toward a second, stationary proton. What happens as the<br> protons get closer?
Vlad [161]

Answer:

they repel

Explanation:

because like charges repel

5 0
3 years ago
Read 2 more answers
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