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Vinil7 [7]
3 years ago
10

How are half life and radioactive decay related

Chemistry
1 answer:
hoa [83]3 years ago
3 0

Answer : Half life and radioactive decay are inversely proportional to each other.

Explanation :

The mathematic relationship between the half-life and radioactive decay :

N=N_oe^{-\lambda t}              ................(1)

where,

N = number of radioactive atoms at time, t

N_o = number of radioactive atoms at the beginning when time is zero

e = Euler's constant = 2.17828

t = time

\lambda = decay rate

when t=t_{1/2} then the number of radioactive decay become half of the initial decay atom i.e N=\frac{N_o}{2}.

Now substituting these conditions in above equation (1), we get

\frac{N_o}{2}=N_oe^{-\lambda t_{1/2}}

By rearranging the terms, we get

\frac{1}{2}=e^{-\lambda t_{1/2}}

Now taking natural log on both side,

ln(\frac{1}{2})=-\lambda \times t_{1/2}

By rearranging the terms, we get

t_{1/2}=\frac{0.693}{\lambda}

This is the relationship between the half-life and radioactive decay.

Hence, from this we conclude that the Half life and radioactive decay are inversely proportional to each other. That means faster the decay, shorter the half-life.

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Alenkinab [10]

<u>Answer:</u> The spacing between the crystal planes is 4.07\times 10^{-10}m

<u>Explanation:</u>

To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

n\lambda =2d\sin \theta

where,

n = order of diffraction = 2

\lambda = wavelength of the light = 154pm=1.54\times 10^{-10}m     (Conversion factor:  1m=10^{12}pm )

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\theta = angle of diffraction = 22.20°

Putting values in above equation, we get:

2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m

Hence, the spacing between the crystal planes is 4.07\times 10^{-10}m

4 0
3 years ago
Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

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3) The reaction must:

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4) The concentration of N2 will:

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Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

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4) The concentration of N2 will:

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Best regards.

8 0
3 years ago
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Maslowich

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Explanation:

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Luba_88 [7]

Answer:

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Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

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^         ^

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-------------------  x  ---------------------  x  --------------------  =  223 g O₂
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