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2 years ago

An isotope has a half-life of 10 minutes. after 20 minutes, what percentage of the original nuclei remain?

1 answer:

Elena-2011 [213]2 years ago

6 0

To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ

where, t1/2 is half-life and λ is the decay constant.

t1/2 = 10 min = 0.693 / λ

Hence, λ = 0.693 / 10 min - (1)

Nt = Nο e∧(-λt)

Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken

by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)

From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500

Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%

Hence, Percentage of remaining nuclei is 25.00%

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If metal ions in a solution were reduced, what would you expect to see?

marysya [2.9K]

Answer:

Solid metal

Explanation:

The reduced form of metal ions is the metal in elemental state (simple substance). So, if you have a solution with metal ions and they are reduced, you probably will see the deposition of the metal. For example: if you have a solution with sodium ions (Na⁺), and the ions are then reduced, you will see the aparition of a solid phase of metallic sodium (Na(s)), according to the following half-reaction:

Na⁺ + e- → Na(s)

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2 years ago

What is the formula for frequency?

abruzzese [7]

Answer:

Frequency = Velocity / Wavelength

f = v / λ

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3 years ago

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago

A fixed mass of gas has a volume of 92 cm cube and 3 degrees Celsius. What will be its volume at 18 degrees celsius if the press

8090 [49]

Answer:

94.8454

Explanation:

Let volume be V

Let Temperature be T

V1= 92

T1= 3C but to kelvin 273+3= 300K