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Serga [27]
2 years ago
13

Six Carnot engines operating between different hot and cold reservoirs are described below. The heat energy transferred to the g

as during the isothermal expansion phase of each cycle is indicated.
Chemistry
1 answer:
Alla [95]2 years ago
7 0

Answer:

<u>Part -A</u>

The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

<u>Part-B</u>

Change in entropy over a complete cycle is 1 =2 = 3= 4  = 5 = 6

Explanation:

<u>Part -A</u>

The change in entropy for the reversible process that transfers the heat energy 'Q' at temperature 'T' is

\Delta S = \frac{Q}{T}

1)

\Delta S_{1} = \frac{Q}{T}

= \frac{1000J}{600K}= 1.67J/K

2)

\Delta S_{2} = \frac{Q}{T}

= \frac{1000J}{500K}= 2J/K

3)

\Delta S_{3} = \frac{Q}{T}

= \frac{1500J}{600K}= 2.5J/K

4)

\Delta S_{4} = \frac{Q}{T}

= \frac{1000J}{500K}= 2J/K

5)

\Delta S_{5} = \frac{Q}{T}

= \frac{500J}{500K}= 1J/K

6)

\Delta S_{6} = \frac{Q}{T}

= \frac{500J}{600K}= 0.83J/K

Therefore, The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

<u>Part-B</u>

\Delta S = \frac{Q}{T}

Q_{system}=work\,\,done

Initial and final states of complete cycle are equal. after complete the one cycle the system returns to the original state. So, the change in the entropy will be the zero

Therefore, Change in the overall cycle is 1 =2 = 3= 4  =5=6

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What mass of ag2co3 could you produce from 12.7 g agno3 assuming that it is the limiting reagent?
zhenek [66]

10.3 g of  Ag_{2} CO_{3} will produce from 12.7 g AgNO_{3}.

In simple terms, a limiting reagent is a reactant that is completely used up in the reaction. 

It is also referred to as a limiting reactant or limiting agent.

Now, according to the question,

Since it is a limiting reagent, it will react fully. 

The mass of AgCO_{3} that will be produced will be:-

Equivalent mole of AgNO_{3} = Equivalent weight of Ag_{2} CO_{3} = 12.7/169.97 = x*2/274x = 10.3 g. 

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It is used to restrict the reaction.

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7 0
1 year ago
What is the empirical formula for Hg2(NO3)2
den301095 [7]
<span>Answer: HgNO₃
</span><span />

<span>Explanation:
</span>
<span /><span /><span>
The empirical formula is the formula that shows the ratio of the atoms in its simplest form, this is using the smallest whole numbers.
</span><span />

<span>The empirical formula may or may not be the same molecular formula.
</span><span />

<span>In this case you are given the molecular formula Hg₂(NO₃)₂. Since, the ratio of the atoms of Hg, N, and O is 2: 2: 6, respectively, the same ratio is expressed if you divide by the greatest common factor (GCF).
</span><span>
</span><span>
</span><span>The GCF of 2, 2, and 6 is 2. So, the ratios can be simplified to 1:1:3, meaning 1 mol of Hg, 1 mol of N, and 3 mol of O or HgNO₃.</span>
8 0
3 years ago
If the average atomic mass of hydrogen in nature is 1.0079, what does that tell you about the percent composition of H-1 and H-2
vovangra [49]

Answer:

That the isotope H-1 is the most abundant in nature.

Explanation:

Hello!

In this case, since the average atomic mass of an element is computed considering the mass of each isotope and the percent abundance each, for hydrogen we would set up something like this:

m_H=m_{H_1}*\%abund_{H_1}+m_{H_2}*\%abund_{H_2}

Moreover, since the isotope notation H-1 and H-2 means that the atomic mass of H-1 is 1 amu, that of H-2 is 2 amu and the average one is 1.0079 amu, we can infer that the most of the hydrogen in nature is H-1 as the most of it composes the average hydrogen atom.

Best regards!

4 0
2 years ago
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