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timama [110]
3 years ago
7

Arrange these compounds by their expected vapor pressure: H2O NCl3 Br2

Chemistry
1 answer:
Tamiku [17]3 years ago
6 0

Answer: Given compounds are arranged in decreasing order of their vapor pressure as follows.

        Br_{2} > NCl_{3} > H_{2}O

Explanation:

Vapor pressure is defined as the pressure exerted by the vapors which are present on the surface of a liquid. Less is the intermoecular forces present within the molecules of a substance more will be its vapor pressure.

For example, Br_{2} is a gas and there exists Vander waal forces between its molecules. Hence, it has more vapor pressure than NCl_{3} and H_{2}O.

In H_{2}O, there exists hydrogen bonding which is stronger than bonding present in NCl_{3}. So, H_{2}O has low vapor pressure.

In NCl_{3}, there exists dipole-dipole interaction which is stronger than Vander waal forces but weaker than hydrogen bonding.

Therefore, given compounds are arranged in decreasing order of their vapor pressure as follows.

        Br_{2} > NCl_{3} > H_{2}O

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An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin
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Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

2NH_3(aq)+CO_2 →CH_4N_2O(aq)+H_2O (l)

Molar mass of urea CH_4N_2O= 60g/mole

Moles of NH_3 = \frac{62.8kg/mole}{17g/mole} (since Moles=\frac{mass  of  substance}{mass of one mole})

                     = 4011.76 moles

Moles of CO_2 = \frac{105kg}{44g/mole}

                = \frac{105000g}{44g/mole}

                = 2386.36 moles

Theoritically , moles of NH_3 required = double the moles of CO_2

    but , 4011.76 , the limiting reagent is NH_3

Theoritical moles of urea obtained = \frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3

                                                      = 2005.88mole CH_4N_2O

Mass of 2005.88 mole of CH_4N_2O =2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}

                                                     = 120352.8g

                                                     120352.8g\times \frac{1kg}{1000g}

                                                     = 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = \frac{87.5kg}{120.35kg}\times100

                                 72.70%

Thus , the percent yeild for the reaction is 72.70%

8 0
3 years ago
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