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fomenos
2 years ago
7

In a heat engine, 700 J of heat enters the system, and the piston does 400 J of work. What is the final internal (thermal) energ

y of the system if the initial energy is 1200 J? 300 J 900 J 1100 J 1500 J.
Chemistry
2 answers:
MAVERICK [17]2 years ago
6 0

Answer: The answer would be: 1500 J

Explanation:

took the quiz

Vitek1552 [10]2 years ago
4 0

The first law of thermodynamics characterises the two types of energy transfer, as heat and as thermodynamic. The final internal (thermal) energy of the system is 1,500 J.

<h3>What is internal energy?</h3>

The energy present in a system itself for conducting reactions is called internal energy.

Given,

  • Heat entering system (Q) = 700 J
  • Work done by the piston (W) = 400
  • Initial energy (\rm U_{1})= 1200 J

According to the <u>first law of thermodynamics</u>:

\rm Q = \Delta U + W

Substituting values in the above equation:

\begin{aligned}\rm Q &= \rm U_{2} - U_{1} + W\\\\\rm U_{2}  &= \rm Q - W + U_{1}\\\\\rm U_{2}  &= 700 - 400 + 1200\\\\&= 1500 \;\rm J\end{aligned}

Therefore, option D. 1500 J is the final energy.

Learn more about internal energy here:

brainly.com/question/2602565

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
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Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

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                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

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(b)   No. of moles of HCl added = Molarity \times volume

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                                             = 0.0116 mol

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Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

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            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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