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rewona [7]
3 years ago
11

Which type of radioactive decay only releases energy?

Chemistry
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

A gamma rays

Explanation:

Gamma rays unlike beta and alpha particles releases energy and they are actual rays and not particles.

A gamma ray is  form of electromagnetic radiation with no mass and charge. Its energy could range from 10KeV to 3MeV.

Its range is not definite in any medium but much longer than those of alpha and beta particles. Intensity of the rays diminishes exponentially from the source.

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At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressur
fenix001 [56]

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the  partial pressure of He would give a solubility of 0.200 M.

<h3>What is Henry's Law ?</h3>

Henry's Law is a gas law states that at a constant temperature the amount of gas that dissolved in a liquid is directly proportional to the partial pressure of that gas.

<h3>What is relationship between Henry's Law constant and Solubility ?</h3>

The solubility of gas is directly proportional to partial pressure.

It is expressed as:

S_{\text{gas}} = K_{H} P_{\text{gas}}

where,

S_{\text{gas}} = Solubility of gas

K_{H} = Henry's Law constant

P_{\text{gas}} = Partial pressure of gas

Now put the values in above expression we get

S_{\text{gas}} = K_{H} P_{\text{gas}}

0.080M = K_{H} × 1.7 atm

K_{H} = \frac{0.080\ M}{1.7\ \text{atm}}

      = 0.047 M/atm

Now we have to find the partial pressure of He

S_{\text{gas}} = K_{H} P_{\text{gas}}

0.200 M = 0.047 M/atm × P_{\text{gas}}

P_{\text{gas}} = \frac{0.200 M}{0.047\ \text{M/atm}}

       = 4.25 atm

Thus from the above conclusion we can say that At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. 4.25 atm is the  partial pressure of He would give a solubility of 0.200 M.

Learn more about the Henry's Law here: brainly.com/question/23204201

#SPJ4

3 0
1 year ago
Are parts of the circulatory system<br> .The<br> , and
Sphinxa [80]

Answer:

The heart and the blood vessels are a part of the circulatory system. The blood vessels include the arteries, veins and capillaries. The lungs are considered to be the pulmonary part of the circulatory system. The heart is the cardiovascular part of the circulatory system and the vessels are the systemic part of the circulatory system. The main function of the circulatory system is to supply all parts of the body with oxygenated blood and to take away the deoxygenated blood from all parts of the body.

8 0
3 years ago
Convert 380 mmHg to atm.
tankabanditka [31]

Answer:

50662.5

Explanation:

6 0
2 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
What is the Relative Formula Mass of Potassium bromide?
Aleonysh [2.5K]
The relative formua for Potassium bromide is Kbr
6 0
3 years ago
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