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Murrr4er [49]
3 years ago
15

After each half-life cycle, of the radioactive nuclei became nonradioactive.

Chemistry
1 answer:
kherson [118]3 years ago
7 0

Answer:

half

Explanation:

i just took the quiz

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Oxidation number of au in kaucl4
DanielleElmas [232]
KauCl4 :

K = + 1 

au = + 7

Cl = - 2

hope this helps!

3 0
3 years ago
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Why do a car's tires need more air in the winter than they do in the summer?
Hunter-Best [27]

Answer: Option (d) is the correct answer.

Explanation:

In winter's, temperature of atmosphere is low and due to this molecules of air present in the tire come closer to each other as they gain potential energy and loses kinetic energy.

Hence, air pressure decreases and there is need to fill more air in the tire.

Whereas is summer's, temperature is high so, molecules of air inside the tire gain kinetic energy and move rapidly from one place to another due to number of collisions. So, air pressure increases and there is no need to fill more air inside the tire.

Thus, we can conclude that the temperature is lower, so the air inside the tires contracts.

3 0
3 years ago
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Which factor affects a solute's solubility rather than its rate of solution?
Illusion [34]

Answer: A

Explanation:

I took the test it was A

4 0
3 years ago
Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles
Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃  +  O₂  ⟶  N₂  +  H₂O

The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0
3 years ago
For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
  • brainly.com/question/19671384
5 0
3 years ago
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