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Sveta_85 [38]
3 years ago
10

Two aqueous solutions are prepared: 1.00 m Na2CO3 and 1.00 m LiCl. Which of the following statements is true?

Chemistry
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

The Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

Explanation:

As concentrations of two aqueous solutions are same therefore we can write:

                                \Delta P\propto i , \Delta T_{b}\propto i and \pi \propto i

where \Delta P, \Delta T_{b} and \pi are lowering of vapor pressure, elevation in boiling point and osmotic pressure of solution respectively. i is van't hoff factor.

i = total number of ions generated from dissolution of one molecule of a substance (for strong electrolyte).

Here both Na_{2}CO_{3} and LiCl are strong electrolytes.

So, i(Na_{2}CO_{3})=3 and i(LiCl)=2

Hence, lowering of vapor pressure, elevation in boiling point and osmotic pressure will be higher for Na_{2}CO_{3} solution.

Therefore the Na_{2}CO_{3} solution has a higher osmotic pressure and higher boiling point than LiCl solution.

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According to Newton’s second law of motion, when an object is acted on by an unbalanced force, how will that object respond? It
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C) It will accelerate.  

Explanation:

According to Newton’s second law of motion, when an object is acted on by an unbalanced force, it will accelerate.

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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
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