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Taya2010 [7]
2 years ago
14

Please For a brainlist and +10 points

Chemistry
1 answer:
denis23 [38]2 years ago
6 0

Answer:

edfgkvisiaixiicciciviicsiaiaqwododc

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1.
Serhud [2]

Explanation:

Given that,

The frequency of electromagnetic spectrum is 2.73\times 10^{16}\ Hz

(A) Let the wavelength of this radiation is \lambda. We know that,

c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{2.73\times 10^{16}}\\\\\lambda=1.09\times 10^{-8}\ m

So, the wavelength of this radiation is 1.09\times 10^{-8}\ m.

(B) Let E is the energy associated with this radiation. Energy of an electromagnetic radiation is given by :

E=hf

h is Planck's constant

E=6.63\times 10^{-34}\times 2.73\times 10^{16}\\\\E=1.8\times 10^{-17}\ J

1 kcal = 4184 J

It means,

1.8\times 10^{-17}\ J=\dfrac{1}{4184}\times 1.8\times 10^{-17}\\\\=4.3\times 10^{-21}\ \text{kcal}

Hence, this is the required solution.

3 0
3 years ago
If the percent transmittance is is 64.9 in 0.2M standard CuSO4, what is the absorption?
Natali [406]
Explain it a little more ?
5 0
3 years ago
What would happen to a sealed bag of chips left in the sun?
olchik [2.2K]

The answer is the gaz inside the bag would expand. (A)

7 0
3 years ago
Read 2 more answers
what conclusions can be made about the relationship between metallic character and the atomic radius?
kolezko [41]

We have to get the relationship between metallic character and atomic radius.

Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.

If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.

With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.

With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.

5 0
2 years ago
Phosphorus-32 is radioactive and has a half life of 14.3 days. What percentage of a sample would be left after 12.6 days
nadya68 [22]

Answer:

There is 54.29 % sample left after 12.6 days

Explanation:

Step 1: Data given

Half life time = 14.3 days

Time left = 12.6 days

Suppose the original amount is 100.00 grams

Step 2: Calculate the percentage left

X = 100 / 2^n

⇒ with X = The amount of sample after 12.6 days

⇒ with n = (time passed / half-life time) = (12.6/14.3)

X = 100 / 2^(12.6/14.3)

X = 54.29

There is 54.29 % sample left after 12.6 days

8 0
2 years ago
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