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notsponge [240]
3 years ago
5

What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same t

ime. 1 cal = 4.184 J.
Chemistry
1 answer:
kenny6666 [7]3 years ago
4 0

Answer: 20.7 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system=-P\Delta V  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = 4.51\times 4.184kJ=18.9kJ    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

\Delta E=+1.79+(18.9)=20.7kJ

Thus \Delta E for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ

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Answer:

1 million years

Explanation:

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The half life of a radioactive isotope is independent of the amount of starting material. Hence, whether the amount of starting material is small or large, the half life of the substance remains the same.

Hence the half life of 2 green marbles and 500,000 green marbles is 1 million years.

8 0
2 years ago
20 ml of a solution of sucrose contains 850 mg of sucrose. What is the weight/volume percentage concentration of this solution i
MissTica
<h3>Answer:  4.25 g/ml %</h3>

Explanation:

weight/volume percentage concentration = (mass in g  ÷  volume) × 100

                                                                      =  (0.850 g ÷ 20 ml) × 100

                                                                      = 4.25 g/ml %

∴ the weight/volume percentage concentration of the sucrose solution is 4.25 g/ml %.

7 0
3 years ago
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Answer:

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(Photo for proof at the bottom.)

Explanation:

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Here's a photo of the unit review on Edge. Refer to the 2nd attachment for a visualization.

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