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Free_Kalibri [48]
2 years ago
9

If a 600 gram radioactive sample goes through 3 half-lives, how would be left?

Chemistry
1 answer:
olga_2 [115]2 years ago
5 0

Answer:

75 g

Explanation:

1 half life = 600/2 = 300g

2 half life = 300/2 = 150g

3 half life = 150/2 = 75g

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Can someone explain this to me, please?
zimovet [89]

Answer:

Isotopes have same atomic numbers, no. of protons and no. of electrons. Only their no. of neutrons and atomic mass are changed.

<u>Na - 24:</u>

Atomic Mass = 24

Atomic No. = 11

Hence,

No. of protons in Na-24 = 11

No. of neutrons = Atomic Mass - Atomic Number

No. of neutrons = 24 - 11

No. of neutrons = 13

Atomic Number = 11

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>
5 0
2 years ago
What controls traits in organisms? A.Cell Membrane B.Cell Wall C.Genes D.Punnett Square
BARSIC [14]

Answer:

I believe it's C (the genes)

3 0
3 years ago
Read 2 more answers
What are two things that you observed when you heated the mixture of tin and nitric acid over the Bunsen burner in the virtual l
Rus_ich [418]
1. The reaction for this would be:

Sn + 4 HNO₃ →  SnO₂ + 4 NO₂ + 2 H₂O

The first observation would be bubbling of the solution and brown acrid smoke is produced due to the presence of NO₂ gas. Another observation would be the presence of a white solid which is SnO₂.

2. Heating was required to get rid of the H₂O. When all moisture is gone, you weigh the sample. Afterwhich, you further heat it to get ride of the oxygen. By doing this, you would know the individual mass of each element. Then, you can solve for the empirical formula of the oxide of tin.
3 0
3 years ago
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a
Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})

Where:

Q: is the energy transferred = 5.0 MJ

k_{B}: is the Boltzmann constant = 1.38x10⁻²³ J/K  

T_{i}: is the initial temperature = 1000 K

T_{f}: is the final temperature = 500 K

Hence, the entropy change is:

\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}                                    

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!          

7 0
3 years ago
He reaction produces 1.6 g of gas in 30 seconds.Calculate the mean rate of the reaction in the first 30 seconds.Use the equation
devlian [24]

Answer:

Mean rate of reaction produced = 0.533 g/sec (approx.)

Explanation:

Given:

Reaction produced = 1.6 gram

Time taken = 30 sec

Find:

Mean rate of reaction produced

Computation:

Mean rate of reaction produced = Reaction produced / Time taken

Mean rate of reaction produced = 1.6 / 30

Mean rate of reaction produced = 0.533 g/sec (approx.)

5 0
3 years ago
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