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galina1969 [7]
3 years ago
5

The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
3 0

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

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Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

Explanation:

The relation between energy and wavelength is as follows.

E = \frac{hc}{\lambda}\\

This means that energy is inversely proportional to wavelength.

As it is given that energy of a hydrogen atom and a helium atom is same.

Let us assume that E_{hydrogen} = E_{helium} = E'. Hence, relation between their wavelengths will be calculated as follows.

E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}    ... (1)

E_{helium} = \frac{hc}{\lambda_{helium}}         ... (2)

Equating the equations (1) and (2) as follows.

E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'

Thus, we can conclude that if a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.

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3 years ago
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3 years ago
23) What is the volume of 2.454 x 1024 atoms of nitrogen gas at STP?
earnstyle [38]

Answer:

d) V =  91.3 L

Explanation:

Given data:

Volume of nitrogen = ?

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Pressure = standard = 1 atm

Number of atoms of nitrogen = 2.454×10²⁴ atoms

Solution:

First of all we will calculate the number of moles of nitrogen by using Avogadro number.

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V =  91.3 atm.L /1 atm

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