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DaniilM [7]
2 years ago
7

7. Suppose each of these isotopes emits a beta particle. Give the iso-

Chemistry
1 answer:
Volgvan2 years ago
7 0

In a beta emission, the mass number of the daughter nucleus remains unchanged while the atomic number of the daughter nucleus increases by one unit. The following are isotopes produced when the following undergo beta emission;

1) potassium-42  ------> Ca - 42

2) iodine-131 ------------> Xe - 131

3)  iron-52 ---------------> Co - 52

4)  sodium-24 -----------> Mg -24

The daughter nucleus formed after beta emission is found one place after its parent in the periodic table.

Regarding the stability of the daughter nuclei, a nucleus is unstable if the neutron-proton ratio is less than 1 or greater than 1.5.

Hence, the following daughter nuclei are stable; Ca - 42, Xe - 131, Mg -24.

Learn more: brainly.com/question/1371390

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Answer:

I think everything that's there is correct.

Explanation:

Bleach with pH of about 12.5 would be a strong base.

Milk with pH of 6.5 and Blood with pH around 7 would be neutral.

Battery acid with pH below 1 is strong acid and orange juice is weak acid.

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Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

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