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Anvisha [2.4K]
3 years ago
7

How much of a radioactive parent isotope will remain after three half-lives have passed?

Chemistry
1 answer:
Sedaia [141]3 years ago
4 0
After three half lives have passed, there would be only 12.5 percent of the original amount of a radioactive parent isotope that will remain. Half life is the time needed for a certain amount of a substance to be half its initial amount. It is a common term used in nuclear chemistry describing how fast radioactive substances undergo decay. One half life would correspond to only 50% would be left. Two half lives would be 25% only of the original value. Three half lives would be 12.5%. Four half lives would be 6.25% of the initial value. So on and so forth.
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2NaCl → 2Na +Cl2<br> What reaction is this
babunello [35]

Answer:

Decomposition reaction

Explanation:

A single reactant breaking down to form 2 or more products is decomposition

5 0
3 years ago
Salt domes result when: 1. inland seas dry 2.the pressure of overlying rock forces the salt to rise 3. a layer of sedimentary sa
Ilia_Sergeevich [38]

Answer:

Salt domes result when <u><em>the pressure of overlying rock forces the salt to rise. (Option 2)</em></u>

Explanation:

In geology it is called the gently wavy and rounded relief dome.

Salt has some special properties like rock:

  • Salt has a lower specific gravity in relation to a common mineral.
  • Salts deform plastically and are very mobile.
  • Salts have a high water solubility.

These properties allow, if the pressure is very high, that the salt layers move upwards (due to their lower density). That is, the internal forces produce the elevation of the strata by means of the pressure they exert towards a higher point, generating that the salt looks for its way towards the surface [that is, the salt ascends through the sedimentary layers of the earth's crust, crossing them and deforming them] and causing the bulging structure. The oldest strata are located in the central area of the dome, while the most modern are distributed in the farthest radius. The structure is called salt or diapiro dome, the phenomenon by which it is formed is called diapirism.

Finally, you can say that <u><em>Salt domes result when the pressure of overlying rock forces the salt to rise.</em></u>

3 0
4 years ago
Read 2 more answers
PLEASE HELP
AfilCa [17]
I think it’s salt water
7 0
3 years ago
Read 2 more answers
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
The bond type in CaO is
sergey [27]
CaO is an ionic bond
4 0
3 years ago
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