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3 years ago

14

What electron transition represents a gain of energy?

1 answer:

leva [86]3 years ago

4 0

Answer:

between the 2 and the 3 shell

Explanation:

when a gain of energy is the displacement of the electrons in a layer of low-energy shell high-energy, and here we have the 2nd shell is the shell of low energy, and 3 shell the shell is high-energy.

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The sum of all chemical reactions that occur in a cell is called:.

vodomira [7]

Answer:

Metabolism

Explanation:

The sum of the chemical reactions that take place within each cell of a living organism and that provide energy for vital processes and for synthesizing new organic material.

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2 years ago

What tool do you use to measure mass in chemistry?

Kryger [21]

Answer:

Balance Mass

Explanation:

It could be Balance Mass

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3 years ago

Suppose 0.09886 M KOH is titrated into 15.00 mL H2SO4 of unknown concentration until the equivalence point is reached. It takes

Mamont248 [21]

Answer:0.0014458

That’s the right one

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3 years ago

Read 2 more answers

I am very confused, our teacher didn’t properly teach us this. How do you do it?

Fudgin [204]

Answer:

You multiply the 3 numbers together to get your volume, in this case it would be 4058.488 cm^3 (cm cubed)

so

V: 4058.488cm^3 ( round up to 4058.5 for convenience)

M: 27579

D: ?

So we divide mass by the volume to get density, which is

27579 / 4058.5 = ~6.79536 (can round up to 7 or 6.8)

This graph can help a lot, so maybe try and memorize it, hope I helped.

3 0

2 years ago

Read 2 more answers

A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.

Dmitrij [34]

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, $0.00388Ms^{-1}$

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

$t_{1/2}\propto \frac{1}{[A_o]^{n-1}}$

or,

$\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}$

or,

$n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1$ .............(1)

where,

$t_{1/2}$ = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

$n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1$

$n=0.000196\approx 0$

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

$t_{1/2}=\frac{[A_o]}{2k}$

When,

$t_{1/2}$ = 29.5 s

$[A_o]$ = 0.229 M

$29.5s=\frac{0.229M}{2k}$

$k=0.00388Ms^{-1}$

Thus, the value of rate constant is, $0.00388Ms^{-1}$

4 0

3 years ago