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lilavasa [31]
2 years ago
8

14.americium- 241 is widely used in smoke detectors. the radiation released by this element ionizes particles that are then dete

cted by a charged particle collector. the half-life of am-241 is 432.2 years, and it decays by emitting alpha particles. how many alpha particles are emitted each second by a 5.00 g sample of am-241
Chemistry
1 answer:
avanturin [10]2 years ago
5 0

Hence, 5.1 * 10^-11 alpha particles is being emitted each second by the sample.

<h3>What is half life?</h3>

The term half life is the time taken for only half of the original  amount of radioactive nuclides to remain.

Thus, we have that the half life is 432.2 years or 1.36 * 10^10 s.

k = 0.693/ 1.36 * 10^10 s

k = 5.1 * 10^-11 alpha particles per second.

Learn more about alpha particles:brainly.com/question/2288334

#SPJ1

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Of the bonds c-c, cc, and c≡c, the c-c bond is ___
Mkey [24]
 The line in between indicates that there is a single bond connecting the two carbon atoms. Atoms can be bonded via single bond, double bond and a triple bond. Among the three, the single bond is the weakest, but it allows for more atoms to connect to it because it leaves 3 more pairs available for bonding.

So in summary, among all the bonds that was given, a c-c bond is the weakest and the longest.
4 0
4 years ago
In 1676, Robert Plot discovered a large bone in a quarry in Cornwall, England. He observed that the bone looked similar to a hum
sergeinik [125]

Answer:

D

Explanation:

4 0
3 years ago
How many grams of glucose (C6H12O6) can be formed from 22 grams of CO2?
Shkiper50 [21]
The catabolism of glucose has an equation of C6H12O6 + 6O2 = 6CO2 +6 H20. Hence for every mole of glucose, 6 moles of CO2 is produced. Given 22 grams of CO2, that is 0.5 mol CO2, we multiply this by 1/6, we get the number of moles of glucose equal to 1/12 mol. The mass of glucose needed is obtained by multiplying this by molar mass of glucose which is 180 g/mol. This is equivalent 15 grams of glucose.
8 0
3 years ago
You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
telo118 [61]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

8 0
4 years ago
Predict which of the following changes will cause the volume of the balloon to increase or decrease assuming that the temperatur
IrinaVladis [17]

Answer:

1. a

2. a

3. b

4. c

Explanation:

The STP is the Standard conditions of Temperature and Pressure, in these conditions, T = 0°C and P = 1 atm. By the ideal gas law,

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We can observe that pressure and volume are indirect proportional, so when one increase, the other decrease. so:

1. The pressure decreases from 1.15 atm to STP, 1 atm, thus the volume increases.

2. The pressure decreases from STP, 1 atm, to 0.5 atm, thus the volume increases.

3. The pressure increases from STP, 1 atm, to 1.25 atm, thus the volume decreases.

4. The pressure remains constant at 1 atm, thus the volume is unchanged.

7 0
3 years ago
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