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labwork [276]
3 years ago
7

Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. which con

tainer has greater radioactive decay?The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.
Chemistry
1 answer:
enyata [817]3 years ago
8 0

Answer:

The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.

Explanation:

From the question,

Container A contains 1000 atoms

Container B contains 500 atoms

<u>The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.</u>

The reason for such is due to the difference in the concentration of the isotopes. Container A which contains higher number of atoms will have the more changes of the release of the neutron as the changes of the hitting and splitting increases as the density of the atoms increases.

<u>Thus, the atoms in the container A will therefore decay faster than the atoms in the container B. </u>

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3. 200 ml of 0.2 M HCl is neutralized with 0.1 M
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Answer:

  • <u><em>a. 0.1 M</em></u>

Explanation:

By definition, <em>half neutralization</em> is the point at which half of the acid has been neutralized.

The neutralization reaction that you are studying is the acid-base reaction:

  • HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)

Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:

  • Half molarity = M / 2 = 0.2 M / 2 = 0.1 M

So, the answer is the first choice: a. 0.1 M.

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In distillation,separation is achieved by difference in---​
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relative volatility.

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Predict the outcome if a solution of yellow
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Answer: The solution will remain yellow.

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Which of the following is a physical property of a substance?? A. Combustion B. Corrosion C. Tendency to tarnish D. Solubility
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A physical property does not change the substance. 

Solubility would be the answer since all of the rest are changing the substance. They all deal with bonds except solubility. 

Answer: D. Solubility

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A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
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(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

6 0
3 years ago
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