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Amiraneli [1.4K]
3 years ago
12

Alpha-emitting substances, such as radon gas, can be a serious health hazard only if A. your eyes are open. B. their radiation s

trikes the skin. C. exposure to them is external. D. none of the above
Chemistry
2 answers:
kipiarov [429]3 years ago
5 0

Alpha-emitting substances, such as radon gas, can be a serious health hazard only if \boxed{{\text{their radiation strikes the skin}}}.

Further Explanation:

When an unstable nucleus tends to stabilize itself by the emission of radiations such as alpha particles, beta particles and gamma rays, the process is called radioactivity. Such substances are known as radioactive substances. This process is also called radioactive disintegration or decay. Uranium, carbon-14, iridium-192 and cobalt-60 are some examples of radioactive substances.

Radon is a radioactive natural gas that is produced from the radioactive disintegration of uranium. It is the second major cause of lung cancer after cigarette smoking. The radioactive particles present in radon gas are released by it when radon is inhaled or ingested in the body. When the skin comes in contact with these radiations, cells and tissues are damaged. It leads to lung cancer. The risk of lung cancer depends on the extent of radon inhalation.

Learn more:

  1. Find the age of the fabric: brainly.com/question/8716136
  2. Which of the following nuclear emissions is negatively charged? brainly.com/question/2552986

Answer details:

Grade: Senior School

Chapter: Nuclear chemistry

Subject: Chemistry

Keywords: radioactivity, radioactive disintegration, uranium, lung cancer, radon, radioactive, radiations, radon inhalation, skin, cigarette smoking, cobalt-60, iridium-192, carbon-14, alpha particles, beta particles, gamma rays, unstable nucleus, emission, radiations.

vodka [1.7K]3 years ago
4 0

Answer:

Explanation:

I picked: A your eyes are open and it was wrong

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
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<span>Temperature and heat energy are closely related.
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A 5 g sample of pure gold has a density of 19.3 g/ml. Your friend purchased a gold ring that was made of 25 g of pure gold. What
patriot [66]

Answer:

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