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Setler [38]
2 years ago
13

How would the law of conservation of energy apply to a backup generator

Chemistry
2 answers:
Karo-lina-s [1.5K]2 years ago
4 0
The law of conservation of energy states that no energy is created nor destroyed. When using a backup generator, the generator uses energy from another resource. 
slavikrds [6]2 years ago
3 0
I'm not trying to statt arguments or anything but this is a physics question... I'm doing it in school
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Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2
erastovalidia [21]

Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4

Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g

6 0
3 years ago
___Al+NaOH=>__Na3AlO3+H2
MaRussiya [10]

Explanation:

<em><u>2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2</u></em>

<em><u>...</u></em>

8 0
3 years ago
How much momentum does a 200 kg rhino have that is running at 35 m/s E
nirvana33 [79]

Answer:

7000 kg*m/s E

Explanation:

Momentum formula: p=mv

m=200kg

v=35 m/s East

p=(200kg)(35m/s E)

m=7000 kg*m/s E

If you want to simplify it further, m=7*10^3 kg*m/s E

5 0
2 years ago
Read 2 more answers
6.02 Ionic and covalent bonds.
Ivan
Convalent Bond energy is low
3 0
3 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
2 years ago
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