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Keith_Richards [23]
3 years ago
7

What is one way that fission and fusion are similar?

Chemistry
2 answers:
nignag [31]3 years ago
8 0
The Correct Answer Is <span>Both Processes Produce Energy From Mass. </span>
Lubov Fominskaja [6]3 years ago
5 0

Explanation:

Fusion reaction is defined as a reaction in which two small nuclei combine together and result in the formation of a large nuclei along with release of huge amount of energy.

For example, ^{2}_{1}H + ^{2}_{1}H \rightarrow ^{4}_{2}He is a nuclear fusion reaction.

On the other hand, a fission reaction is defined as a reaction in which a large nuclei splits into two small nuclei along with release of huge amount of energy.

For example, ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He is a nuclear fission reaction.

Therefore, in both these reactions it is common that energy is releasing due to conversion of mass into energy.

Thus, we can conclude that energy is released by converting mass into energy is one way that fission and fusion are similar.

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What evidence do you see that a chemical reaction is taking place when you add hydrogen peroxide to potassium ioxide??
Yuri [45]

Answer:

Bubbling/Foaming

Heat is produced

Explanation:

A chemical reaction has occurred if

1. There is a change in colour

2. Formation of a precipitate

3. Formation of a gas

4. Change in temperature

5. Change of smell

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3 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

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