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3241004551 [841]
3 years ago
11

When an electron moves to another orbital farther from the nucleus, does it gain or lose energy?

Chemistry
1 answer:
irina [24]3 years ago
3 0
I believe that it loses energy, as it is moving away from the main power/control source.
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How many grams of sodium azide are required to form 16.55g of nitrogen gas in
MrRissso [65]

Answer:

16.8128 grams of sodium azide is needed

Explanation:

2NaN3 = 2Na + 3N2

molar mass of NaN3 =

2(11 + (7  \times 3) = 64

molar mass of N2

3(7 \times 2) = 42

moles of N2 =

moles=mass/R.F.M

16.55÷42 = 0.3940

mole ratio=

2NaN3 : 3N2

3 = 0.3940

2=?

0.394. \times 2 \div 3 = 0.2627

Mass of sodium azide =64×0.2627

=16.8126

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A jet engine might use _____.
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It would use rapid combustion
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What does bacteria cell and plant cell have in common
Natasha_Volkova [10]

Answer:

They both have a cell wall

Explanation:

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NiS2(s) + O2(g) --> NiO(s) + SO2(g) When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. The theoretical
makkiz [27]

Answer:

1. The theoretical yield of NiO is 5.09g.

2. O2 is the limiting reactant.

3. The percentage yield of NiO is 95.5%

Explanation:

Step 1:

The balanced equation for the reaction is given below:

2NiS2(s) + 5O2(g) —> 2NiO(s) + 4SO2(g)

Step 2:

Determination of the masses of NiS2 and O2 that reacted and the mass of NiO produced from the balanced equation. This is illustrated below below:

Molar mass of NiS2 = 59 + (32x2) = 123g/mol

Mass of NiS2 from the balanced equation = 2 x 123 = 246g

Molar mass of o3= 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

Molar mass of NiO = 59 + 16 = 75g/mol

Mass of NiO from the balanced equation = 2 x 75 = 150g

Summary:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2 to produce 150g of NiO

Step 3:

Determination of the limiting reactant. This can be obtain as follow:

From the balanced equation above, 246g of NiS2 reacted with 160g of O2.

Therefore, 11.2g of NiS2 will react with = (11.2 x 160)/246 = 7.28g of O2.

From the above calculation, we can see that it will take a higher mass of O2 i.e 7.28g than what was given i.e 5.43g to react completely with 11.2g of NiS2.

Therefore, O2 is the limiting reactant and NiS2 is the excess reactant.

1. Determination of the theoretical yield of NiO.

In this case, the limiting reactant will be used as all of it is consumed in the reaction. The limiting reactant is O2.

From the balanced equation above, 160g of O2 reacted to produce 150g of NiO.

Therefore, 5.43g of O2 will react to produce = (5.43 x 150)/160 = 5.09g of NiO.

Therefore, the theoretical yield of NiO is 5.09g.

2. The limiting reactant is O2. Please review step 3 above for explanation.

3. Determination of the percentage yield of NiO. This is illustrated below:

Actual yield of NiO = 4.86g

Theoretical yield of NiO = 5.09g

Percentage yield =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 4.86/5.09 x 100

Percentage yield of NiO = 95.5%

3 0
3 years ago
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