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kupik [55]
2 years ago
7

Which descriptions apply to emitted radiation? Check all that apply. measurement in Ci/Bq measurement in conventional rad or the

SI Gy the amount of radioactive materials released into the environment. number of disintegrations of radioactive atoms in a radioactive material over a period of time absorbed dose (the amount of energy deposited per unit of mass of human tissue)
Chemistry
2 answers:
xxTIMURxx [149]2 years ago
7 0
The correct answer is measurement in Ci/Bq, the amount of radioactive materials released into the environment, and <span>number of disintegrations of radioactive atoms in a radioactive material over a period of time. I hope this helps.</span>
Llana [10]2 years ago
6 0

Answer:

- Measurement in Ci/Bq

- The amount of radioactive materials released into the environment.

- Number of disintegrations of radioactive atoms in a radioactive material over a period of time.

Explanation:

Hello,

In this case, the following options are valid:

- Measurement in Ci/Bq : one becquerel is understood as the quantity of a radioactive element in which there is one atomic disintegration per second.

- The amount of radioactive materials released into the environment: this technique is suitable to compute the amount of radioactive materials that are being or were released to the environment.

- Number of disintegrations of radioactive atoms in a radioactive material over a period of time: this technique provides becquerel which were described on the first option.

Best regards.

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13. How many electrons are transferred in the following reaction? (The reaction is unbalanced.) Fe(s) + I'(aq) 12(s) + Fe3+(aq)
Nat2105 [25]

Answer : The correct option is, (C) 6

Explanation :

Oxidation-reduction reaction : It is a reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction : It is the reaction in which a substance looses its electrons. In this oxidation state increases.

Reduction reaction : It is the reaction in which a substance gains electrons. In this oxidation state decreases.

The given unbalanced chemical reaction is,

Fe(s)+I^-(aq)\rightarrow I_2(s)+Fe^{3+}(aq)

Half reactions of oxidation and reduction are :

Oxidation : Fe(s)\rightarrow Fe^{3+}+3e^-        ......(1)

Reduction : 2I^-(aq)+2e^-\rightarrow I_2         .......(2)

In order to balance the electrons, we multiply equation 1 by 2 and equation 2 by 3, we get:

Oxidation : 2Fe(s)\rightarrow 2Fe^{3+}+6e^-        ......(1)

Reduction : 6I^-(aq)+6e^-\rightarrow 3I_2         .......(2)

The overall balanced chemical reaction will be:

2Fe(s)+6I^-(aq)\rightarrow 3I_2(s)+2Fe^{3+}(aq)

From this reaction we conclude that the electrons are getting transferred from iron to iodine and the number of electrons transferred are 6 electrons.

Hence, the correct option is, (C) 6

5 0
2 years ago
What is the molarity of 0.540g Mg N,0. in 250 mL of solution?​
Sati [7]

Answer:

2.16x10^-03

molarity=mass÷volume

0.540÷250

3 0
2 years ago
Which of the following elements could replace copper in a single replacement reaction?
Likurg_2 [28]
I’m thinking it’s gold because lead isn’t with cooper meaning that if u switch lead with cooper it won’t work at all .
7 0
3 years ago
2 Points
barxatty [35]

Answer:

it's food engineering obviously

4 0
3 years ago
Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2
arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

a.

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ :  159.69 g/mol

mol Fe₂O₃

\tt \dfrac{100}{159,69}=0.626

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

\tt \dfrac{2}{1}\times 0.626=1.252

mass of Fe(Ar=55.845 g/mol) :

\tt 1.252\times 55.845=69.92~g

b.

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

\tt =\dfrac{65}{69.92}=0.93=93\%

8 0
3 years ago
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