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harina [27]
2 years ago
10

What is one advantage of developing energy-efficient technologies?

Chemistry
2 answers:
Neporo4naja [7]2 years ago
7 0

Answer:

it's simple

Explanation:

For residential buildings, a whole-house approach using a cost-effectiveness criterion can result in savings of 50 percent or more in heating and cooling and 30–40 percent reductions in total energy use.

I hope you understand have a great day :3

Olegator [25]2 years ago
7 0
One advantage of developing energy-efficient technologies, would be less waisted money on electricity.
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The only nonmetal in Group 14 is<br> A carbon.<br> B. oxygen.<br> C. nitrogen.<br> D. fluorine.
lesantik [10]

Answer:

carbon

Explanation:

i just googled it

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Write a conclusion statement that addresses the following questions:
Sindrei [870]

Answer:

did you ever get the answer lol

6 0
3 years ago
Isotopes of an element have the same number of protons but a different number of<br> d.neutrons.
horsena [70]
Yes this is correct. Change the number of protons and you have a different element
7 0
3 years ago
1.) A gas occupies 3.5L at 2.5 atm pressure. What is the volume at 10 atm pressure?
romanna [79]

Answer:

.875

Explanation:

Use Boyle's Law and rearrange formula.

- Hope this helps! Please let me know if you need further explanation.

5 0
2 years ago
Strotium−90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half-life of 28.1 yr. (a) Calculate
Bess [88]

Answer :

(a) The first-order rate constant for the nuclear decay is, 0.0247\text{ years}^{-1}

(b) The fraction of 90-Sr that remains after 10 half-lives is, \frac{a_o}{1024}

(c) The time passed in years is, 127.4 years.

Explanation :

<u>Part (a) :</u>

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{28.1\text{ years}}

k=0.0247\text{ years}^{-1}

Thus, the first-order rate constant for the nuclear decay is, 0.0247\text{ years}^{-1}

<u>Part (b) :</u>

Now we have to calculate the fraction of 90-Sr that remains after 10 half-lives.

Formula used :

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives

a_o = Initial amount of the reactant

n = number of half lives  = 10

Now put all the given values in the above formula, we get:

a=\frac{a_o}{2^{10}}

a=\frac{a_o}{1024}

Thus, the fraction of 90-Sr that remains after 10 half-lives is, \frac{a_o}{1024}

<u>Part (c) :</u>

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.0247\text{ years}^{-1}

t = time passed by the sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = 100 - 95.7 = 4.3

Now put all the given values in above equation, we get

t=\frac{2.303}{0.0247}\log\frac{100}{4.3}

t=127.4\text{ years}

Therefore, the time passed in years is, 127.4 years.

6 0
3 years ago
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