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Anon25 [30]
4 years ago
5

Chem help please

Chemistry
1 answer:
Murrr4er [49]4 years ago
6 0

Considering the equation:

Li(s)+N₂(g)—->Li₃N(s)

Balancing this equation will give :

6Li(s)+N₂(g)—->2Li₃N(s)

It can be seen that 6 mol of Li reacts with 1 mol of N₂to produce 2 mol of Li₃N.

Here 33.6524g of nitrogen gas reacts with 58.7032g lithium to produce Li₃N.

Moles of Li = 58.7032 g * \frac{1 mol }{6.94 g} = 8.46 mol Li

Moles of N_{2} = 33.6524 g * \frac{1 mol}{28 g} = 1.2 mol N_{2}

Mass of Li_{3} N from Li = 8.46 mol Li * \frac{2 mol Li_{3}N}{6 mol Li}*  \frac{34.83 g Li_{3}N}{1 mol Li_{3}N}  = 98.2 g

Mass of Li_{3} N from [tex] N_{2} = 1.2 mol N_{2} * \frac{2 mol Li_{3}N}{1 mol N2}* \frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 98.2 g [/tex] = 83.6 g

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7. A gas has a volume of 300 mL at 300 mm Hg. What will its volume be if the pressure is changed to 500 mm Hg?​
USPshnik [31]

Answer:

The volume is

<h2>180 mL</h2>

Explanation:

In order to solve for the volume we use the formula for Boyle's law which is

<h3>P _{1}  V _{1} = P _{2}V _{2}</h3>

where

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Since we are finding the final volume we are finding V2

Making V2 the subject we have

<h3>V _{2}  = \frac{P _{1}  V _{1}}{P _{2}  }</h3>

From the question

P1 = 300 mmHg

V1 = 300 mL

P2 = 500 mmHg

Substitute the values into the above formula and solve for the final volume obtained

That's

<h3>V _{2} =  \frac{300 \times 300}{500}  \\  =  \frac{90000}{500}  \\  =  \frac{900}{5}</h3>

We have the final answer as

<h3>180 mL</h3>

Hope this helps you

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