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konstantin123 [22]
3 years ago
5

Consider the three equations below.

Chemistry
1 answer:
mash [69]3 years ago
5 0

Answer:

Nuclear fusion plays an important role in making elements that are heavier than helium.

Explanation:

Nucleosynthesis is the process by which new atomic nuclei are created from pre-existing nucleons (protons and neutrons) and nuclei. According to current theories, the first nuclei were formed a few minutes after the Big Bang, through nuclear reactions in a process called Big Bang nucleosynthesis.

In order to synthesize a new element, there must be a change in the number of protons. We should remember that elements are known by the number of their protons as it represents their atomic number.

Elements heavier than helium are formed by nuclear nucleosynthesis in which nuclear fusion plays a very crucial role as typified by the equations shown in the question.

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PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro
podryga [215]

Answer :

Part 1 : Balanced reaction, 3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

Part 2 : The theoretical yield of NH_3 gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of N_2 = 475 g

Molar mass of N_2 = 28 g/mole

Molar mass of NH_3 = 17 g/mole

Experimental yield of NH_3 = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

<u>Answer for Part (2) :</u>

First we have to calculate the moles of N_2.

\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles

From the given reaction, we conclude that

1 moles of N_2 gas react to give 2 moles of NH_3 gas

16.96 moles of N_2 gas react to give \frac{2}{1}\times 16.96=33.92 moles of NH_3 gas

Now we have to calculate the mass of NH_3 gas.

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

\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g

Therefore, the theoretical yield of NH_3 gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100

\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%

Therefore, the % yield of ammonia is 90.03 %

3 0
3 years ago
Suppose the concentration of the standard stock solution is 50.04 ppm and that you delivered 4.60 mL of this standard stock solu
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Answer:

9.21 ppm

Explanation:

Considering

concentration_{working\ solution}\times Volume_{working\ solution}=concentration_{stock\ solution}\times Volume_{stock\ solution}

Given  that:

concentration_{working\ solution}=?

Volume_{working\ solution}=25.00mL

Volume_{stock\ solution}=4.60mL

concentration_{stock\ solution}=50.04 ppm

So,  

concentration_{working\ solution}\times 25.00mL=50.04 ppm\times 4.60mL

concentration_{working\ solution}=\frac{50.04\times 4.60}{25.00}\ ppm=9.21\ ppm

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Explanation:

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