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lana66690 [7]
3 years ago
14

Gamma radiation:

Chemistry
1 answer:
34kurt3 years ago
3 0

Answer:

All of the above

Explanation:

All given are the properties of gamma radiations. Because,

These can be detected with scientific instruments.

These radiations emitted with alpha or beta emission.

These radiations can penetrate to ma y other materials.

Gamma radiations:

Gamma radiations are high energy radiations having no mass.

These radiations are travel at the speed of light.

Gamma radiations can penetrate into the many materials.

These radiations are also used to treat the cancer.

Lead is used for the protection  against gamma radiations because of its high molecular density.

The lead apron are used by the person when treated with gamma radiations.

Lead shields are also used in the wall, windows and doors of the room where gamma radiations are treated, in-order to protect the surroundings.

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Elias serves a volleyball at a velocity of 16 m/s. The mass of the volleyball is 0.27 kg. What is the height of the volleyball a
frosja888 [35]

Answer:

2,7 m

Explanation:

You can solve this doing an energy balance:

m*g*h-\frac{1}{2} *m*v^{2} =41,7[J]

Solving this equation to get h:

\frac{41,7- \frac{1}{2} *m*v^{2} }{m*g}=h

Replacing the values and solving brings to 2,7 m

6 0
3 years ago
Read 2 more answers
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
Can someone please help me
Lilit [14]
It is A. Barium

Explanation: I did that already
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3 years ago
What volume of H2 at STP, can be released by 42.7 g Zn in the equation: Zn + 2 HCl ————> H2 + ZnCl2
ehidna [41]

Answer:

0.86L.

Explanation:

1mol of Zn has mass of 65.39g . The amount of Zn is 2.5g65.39g/mol=0.038mol

7 0
2 years ago
Consider a transition of the electron in the hydrogen atom from n=5 to n=9. Determine the wavelength of light that is associated
lesantik [10]
Lol i need to answer more and plus n could also = # 4.65
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