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3 years ago

Definition of artficial and natural radioactivity

2 answers:

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The spontaneous emission of radiations from an unstable nuclei is known as natural radioactivity. on the other hand, The process of emission of radiations from naturally occurring isotopes when they are bombarded with sub-atomic particles or high levels of X-rays or gamma rays called artificial radioactivity.

leva [86]3 years ago

3 0

Answer:

The key difference between natural and artificial radioactivity is that natural radioactivity takes place on its own in nature whereas when it is induced by man in laboratories, it is called artificial radioactivity.

Explanation:

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Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in

Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0

3 years ago

using a weston cadmium cell of 1.0283 v and a standard resistance of .1ohm a potentiometer was adjusted so that 1.0183 m was equ

anyanavicka [17]

Answer:

Explanation:

the value is -8 cm

3 0

2 years ago

A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is

Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 × 40 × a = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0

3 years ago

Suppose an automobile has 2000 J of kinetic energy. When it moves at twice the speed, what will be its kinetic energy?

Varvara68 [4.7K]

The answer should be 8000 J

8 0

4 years ago

if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?

LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

The complete question is written below:

In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of $1.80 m/s^{2}$ , it would go from rest to its top speed in 85.6 s.

a) What was the speed of the vessel?

b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute?

Calculate the answers in both meters per second and kilometers per hour

a) The average acceleration $a_{av}$ is expressed as:

$a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t}$ (1)

Where:

$a_{av}=1.80 m/s^{2}$

$\Delta V$ is the variation of velocity in a given time $\Delta t$ , which is the difference between the final velocity $V$ and the initial velocity $V_{o}=0$ (because it starts from rest).