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Sphinxa [80]
3 years ago
11

Which of the following statements describes the correct method of preparation of 1.00 L of a 2.0 M urea solution?

Chemistry
1 answer:
Zolol [24]3 years ago
8 0

Answer:

To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

Explanation:

Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.

So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles

We know, number of moles of a compound is the ratio of mass to molar mass of that compound.

So, mass of  2 moles of urea = (2\times 60.06)g=120 g

Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

So, option (C) is correct.

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Answer:

1. True

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Explanation:

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An endothermic reaction is reaction that will absorb energy from the surrounding arena. Increasing temperature will increase the heat of the system. Since the average heat of the surrounding is higher, it will be easier to do an endothermic reaction than exothermic, so this will shift the equilibrium position toward endothermic reaction.

2.

When pressure increase, the molecule will harder to expand. This mean reaction that produces more molecules will be harder to happen since it will take more room and increase the pressure further. This will make the equilibrium shift toward the side with lower total mole concentration since it will help to make more room, thus making the pressure lower.

3.

Remember that only the gas form will contribute to the pressure of the system. In this reaction, there are 3 kinds of gas: nitrogen, hydrogen, and ammonia. Since all form in this reaction gas, you can calculate them all.

On the left side, we have 1 nitrogen and 3 hydrogen so the total is 4.

On the right side, we have 2 ammonia so the total is 2.

When pressure decrease, the equilibrium will shift toward the side with lower total mole concentration, which is the ammonia side. So, the reaction will shift to the product side.

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