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irakobra [83]
2 years ago
8

the isotope 146c has a half life of 5730 years. what fraction of 146c in a sample with mass ,m, after 28650 years

Chemistry
1 answer:
defon2 years ago
3 0

Answer:

3.1% is the fraction of the sample after 28650 years

Explanation:

The isotope decay follows the equation:

Ln[A] = -kt + Ln[A]₀

<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>

<em />

k could be obtained from Half-Life as follows:

K = Ln 2 / Half-life

K = ln 2 / 5730 years

K = 1.2097x10⁻⁴ years⁻¹

Replacing in isotope decay equation:

Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]

Ln[A] = -3.4657

[A] = 0.0313 =

<h3>3.1% is the fraction of the sample after 28650 years</h3>

<em />

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When of a certain molecular compound X are dissolved in of dibenzyl ether , the freezing point of the solution is measured to be
kondaur [170]

This question is incomplete, the complete question is;

When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.

If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answer: molar mass of solute (X) is 88.03 g/mol

Explanation:

Given that;

mass of solute = 4.28 g

mass of solvent = 60.0 g = 0.060 kg        (Dibenzyl ether)

depression constant kf = 6.17 °CKg/mol

Freezing Point of solvent T₀ = 1.80°C       (Dibenzyl ether)

freezing point of solution Tsol = -3.20°C

Now we know that

Depression in freezing point ΔTf = depression constant kf × molaity m

and (ΔTf = T₀-Tsol)

so T₀ - Tsol = kf × m

we substitute

1.80 - (-3.20) = 6.17  × m

5 = 6.17 × m

m = 5 / 6.17

m = 0.8103 kg/mol

so molaity m = 0.8103 kg/mol

we know that

Molaity of solute m = (mass of solute / M.wt of solute) × ( 1 / mass of solvent in Kg)

solve for molar mass of solute

molar mass of solute =  (mass of solute / molaity) × ( 1 / mass of solvent in Kg)

now we substitute

molar mass = (4.28g / 0.8103 kg/mol) × (1 / 0.060kg)

molar mass = ( 5.2839 × 16.66 ) g/mol

molar mass = 88.0297 g/mol ≈ 88.03 g/mol

Therefore molar mass of solute (X) is 88.03 g/mol

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Explanation:

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What is the mass of 4.5 x 10^22 molecules of hydrogen peroxide (H2O2)? Show your work in the space below.
valentinak56 [21]

Given :

Number of molecules of hydrogen peroxide, N = 4.5 × 10²².

To Find :

The mass of given molecules of hydrogen peroxide.

Solution :

We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.

So, number of moles of hydrogen peroxide is :

n = \dfrac{N}{N_a}\\\\n = \dfrac{4.5\times 10^{22}}{6.022\times 10^{23}}\\\\n = 0.0747 \ moles

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Hence, this is the required solution.

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