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garri49 [273]
3 years ago
14

When do epa’s refrigerant management regulations exempt refrigerant from the venting prohibition

Chemistry
1 answer:
Xelga [282]3 years ago
6 0

The Environmental Protection Agency (EPA) refrigerant management regulations would exempt any refrigerant from venting problem when it determines that the refrigerant in an appliance do not pose a threat to the environment (surrounding) if released.

A refrigerant can be defined as any chemical substance that undergoes a phase change (liquid and gas) so as to enable the cooling and freezing of materials. They are typically used in air conditioners, refrigerators, water dispensers, etc.

In the United States of America, the Environmental Protection Agency (EPA) is a governmental agency which was established by U.S Congress and it is saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, degradation, pesticide use, contamination, and hazardous waste spills. Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act.

Simply stated, the Environmental Protection Agency (EPA) is the governmental agency set up to ensure that various industries, factories and people comply with laws and regulations concerning the environment.

In conclusion, appliances such as a refrigerator, air conditioner (AC), etc., whose refrigerants do not pose a threat (potentially cause damage) to the environment if released are typically exempted by the EPA's refrigerant management regulations.

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Please use the values in the resources listed below instead of the textbook values. Under certain conditions the decomposition o
Alex787 [66]

Answer:

The rate equation for this reaction:

R=k[NH_3]^0

Explanation:

Decomposition of ammonia:

2NH_3\rightarrow N_2+3H_2

Rate law of the can be written as;

R=k[NH_3]^x

1) Rate of the reaction , when [NH_3]=2.0\times 10^{-3} M

1.5\times 10^{-6} M/s=k[2.0\times 10^{-3} M]^x..[1]

2) Rate of the reaction , when [NH_3]=4.0\times 10^{-3} M

1.5\times 10^{-6} M/s=k[4.0\times 10^{-3} M]^x..[2]

[1] ÷ [2]

\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}

On solving for x , we get ;

x = 0

The rate equation for this reaction:

R=k[NH_3]^0

4 0
3 years ago
How does Elnora feel about her future? A) She takes one day at a time. B) She is apprehensive and anxious. C) She is optimistic
Georgia [21]

Answer:

D) she is confident and has already made plans

4 0
3 years ago
What is the strong nuclear force
Arturiano [62]

Answer:

A fundamental interaction of nature that acts between subatomic particles of matter. The strong force binds quarks together in clusters to make more-familiar subatomic particles, such as protons and neutrons. Something like that.

Explanation:

8 0
2 years ago
Sketch two amino acids side-by-side, on one of them label the functional groups, then show how the two can be joined together.
Alinara [238K]

Answer : The sketch of two amino acids side by side are shown below.

Explanation :

Amino acid : Amino acid are the acid that contains two functional groups which are carboxylic group, -COOH and ammine group, -NH_2.

When the two or more that two amino acids join together with the help of peptide bond, they produces polypeptide chain or protein.

The bond present between the two amino acid is called a peptide bond.

The peptide bond is a chemical bond that is formed between the two molecules when the nitrogen of one amino acid react with the carbon of another amino acid by releasing a water molecule.

Hence, the sketch of two amino acids side by side are shown below.

6 0
3 years ago
The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) Th
Fantom [35]

<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant  = 4.82\times 10^{-3}s^{-1}

t = time taken for decay process = 151 sec

[A_o] = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}

[A]=0.041moles

Hence, the amount remained after 151 seconds are 0.041 moles

7 0
3 years ago
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