CHEMISTRY HIGH SCHOOL

For the reaction:6Li(s)+N2(g)→2Li3N(s)
Determine:
the mass of N2 needed to react with 0.536 moles of Li.
the number of moles of Li required to make 46.4 g of Li3N.
the mass in grams of Li3N produced from 3.65 g Li.
the number of moles of lithium needed to react with 7.00 grams of N2.

Answers

Answer 1

Answer:

Explanation:

$6Li(s)+N_2(g)\rightarrow 2Li_3N(s)$

1. Mass of $N_2$ needed to react with 0.536 moles of Li.

According to reaction, 6 moles of Li reacts with 1 mol of $N_2$ .

Then 0.536 moles of Li will react with:

$(1)/(6)* 0.536$ moles of $N_2$ that is 0.0893 moles.

Mass of $N_2[tex] gas needed:</strong></p><p><strong>[tex]=28 g/mol* 0.0893 mol=2.5004 g$

2.The number of moles of Li required to make 46.4 g of $Li_3N$

Moles of $Li_3N=(46.4 g)/(35 g/mol)=1.3257 mol$

According to reaction the 2 moles of $Li_3N$ are produced from 6 moles of Li.

Then 1.3257 moles of $Li_3N$ will produced from:

$(6)/(2)* 1.3257=3.9771 moles$

3.9771 moles of lithium will needed.

3. The mass in grams of $Li_3N$ produced from 3.65 g Li.

Moles of Li $=(3.65 g)/(7 g/mol)=0.5214 moles$

According to reaction, 6 moles of Li gives 2 moles of $Li_3N$

Then 0.5214 moles of Li will give $(2)/(6)* 0.5214$ that is 0.1738 moles of $Li_3N$ .

Mass of $Li_3N=0.1738 mole* 35 g/mol=6.083 g$

6.083 grams of $Li_3N$ will be produced.

4. The number of moles of lithium needed to react with 7.00 grams of $N_2$ .

Moles of $N_2=(7.00 g)/(28 g/mol)=0.25 moles$

1 mol of $N_2$ reacts with 6 mol of Li

Then, 0.25 moles of $N_2$ will ftreact with :

$6* 0.25 moles=1.5 moles$ of lithium

1.5 moles of Li will be needed.

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Answers

An acid base reaction is a netutralization reaction which produces water and salt. It is not a decomposition since the reactants are not broken to simpler elements nor combustion since it does not uses oxygen as reactant.

Answer: Neutralization Reaction

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Answers

The correct answer for the question that is being presented above is this one: "C. solubility level." The term defines the amount that a solute can dissolve into a solution is called the solubility level. The higher the solubility level, the faster it dissolves.

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Answers

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Learn more about the rate of a reaction

brainly.com/question/8592296

D. A ll of the above. catalyst reduces the activation energy thereby making the reaction faster. an increase in temperature increases the kinetic energy and no of collisions making reaction faster for an endothermic reaction while decrease in temperature favors an exothermic reaction.
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Answers

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Sulfur is considered a non-metal. Thanks for posting your question. Feel free to ask me any question. I might not always know the answer. Although i'll try to answer correctly. Have a good day love.

(a) how many milliliters of a stock solution of 6.0 m hno3 would you have to use to prepare 110 ml of 0.500 m hno3? (b) if you dilute 10.0 ml of the stock solution to a final volume of 0.250 l, what will be the concentration of the diluted solution?

Answers

a) in this we are diluting a stock solution, so we can use the dilution formula

c1v1 = c2v2

where c1 is concentration and v1 is volume of the stock solution

c2 is concentration and v2 is volume of the diluted solution to be prepared

substituting the values

6.0 M x V = 0.500 M x 110 mL

V = 9.17 mL

9.17 mL of the stock solution should be taken and diluted upto 110 mL to prepare the 0.500 M solution

In this question we are given the volume taken from the stock solution , we have to find the concentration of the diluted solution

again we use the dilution formula, c1v1 = c2v2

substituting the values

6.0 M x 10.0 mL = C x 250 mL

C = 0.24 M

the concentration of diluted solution is 0.24 M

Final answer:

To prepare 110 ml of 0.500 M HNO3 from a 6.0 M HNO3 solution, 9.17 ml of the stock solution would have to be used. If 10.0 ml of the stock solution is diluted to a final volume of 0.250 L, the concentration of the diluted solution will be 0.24 M.

Explanation:

(a) In order to prepare 110 ml of 0.500 M HNO3 from a 6.0 M HNO3 solution, we have to use the formula M1V1 = M2V2 where M and V are the molarity and volume respectively. Here, the M1 and V1 are the molarity and volume of the stock solution and M2 and V2 are the molarity and volume of the diluted solution. Filling in known values, 6.0M * V1 = 0.500M * 110ml. Solving for V1, we get V1 = (0.500 M * 110 ml) / 6.0 M = 9.17 ml. So, you would have to use 9.17 ml of the stock solution.

(b) The diluted solution's molarity is calculated using the same formula as before. Substituting the known values 6.0M * 10.0 ml = M2 * 0.250 L, rearrange the formula to get M2= (6.0M * 10.0 ml) / 0.250 L = 0.24 M or 240 mM. Therefore, the concentration of the diluted solution is 0.24 M.