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Vaselesa [24]
3 years ago
8

The half-life of Palladium-100 is 4 days. After 12 days a sample of Palladium-100 has been reduced to a mass of 4 mg. What was t

he initial mass (in mg) of the sample?
Chemistry
1 answer:
Degger [83]3 years ago
3 0

Answer : The initial mass of the sample is, 31.9 mg

Explanation :

Half-life = 4 days

First we have to calculate the rate constant, we use the formula :

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

k=\frac{0.693}{4\text{ days}}

k=0.173\text{ days}^{-1}

Now we have to calculate the initial mass of sample.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.173\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 4 mg

Now put all the given values in above equation, we get

12=\frac{2.303}{0.173}\log\frac{a}{4}

a=31.9mg

Therefore, the initial mass of the sample is, 31.9 mg

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3 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

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⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

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∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

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5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

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∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

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⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

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