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3 years ago

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A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life

1 answer:

mixas84 [53]3 years ago

8 0

T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>

<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>

<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>

<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>

<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>

<span>HL = 14.4 s / 2 = 7.2 seconds </span>

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Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of nitrogen gas = $79mL/s$

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$M_1$ = molar mass of nitrogen gas = 28 g/mole

$M_2$ = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}$

$R_2=52mL/s$

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

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3 years ago

Why do you think scientists claim to offer only supporting evidence for a theory and not proof?

larisa [96]

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3 years ago

If the particles of matter that make up a substance are relatively far apart and can move freely, the substance is in what state

Novay_Z [31]

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3 years ago

Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and

tatyana61 [14]

<u>Answer:</u> The equilibrium concentration of $PCl_5$ is 1.285 M.

<u>Explanation:</u>

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The expression for equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

We are given:

$K_c=0.042$

$[PCl_3]=0.18M$

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The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

$0.042=\frac{0.18\times 0.30}{[PCl_5]}$

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Hence, the equilibrium concentration of $PCl_5$ is 1.285 M.

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3 years ago

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2 years ago

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