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daser333 [38]
3 years ago
9

What is one benefit that nuclear power plants currently provide for our environment?

Chemistry
2 answers:
LuckyWell [14K]3 years ago
4 0

Answer:

D

Explanation:

Just did the test

liubo4ka [24]3 years ago
3 0

Explanation:

Nuclear power plants deliver electricity 24 hours a day, 7 days a week, irrespective of weather and seasons. As well as reducing greenhouse gas emissions, nuclear generation helps reduce air pollution

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Which substance has the highest ignition​ temperature?
erica [24]
It will be  a. <span>benzene.</span>
7 0
3 years ago
Select the correct answer.
sladkih [1.3K]
C not sure 100 percent but my best guess
7 0
3 years ago
Ordinary water boils at 100°C. Can it be made to boil at 95°C or 105°C.​
slava [35]

It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.

<h3>Concept :</h3>

To boil water at 95°C, decrease the atmospheric pressure.

At 105°C, the water will be boiling quickly than normal at 100°C.

4 0
2 years ago
Read 2 more answers
Given:
bulgar [2K]

Answer:

The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

Explanation:

Given;

CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol

From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

CH₄ = 12 + (1x4) = 16 g/mol

59.7 g of CH₄ = \frac{59.7}{16} = 3.73125 \ moles

1 mole of methane (CH₄) released 890 kilojoules of energy

3.73125 moles of methane (CH₄) will release ?

= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
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