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2 years ago

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

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

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>

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mel-nik [20]

Answer:

The correct answer is b.

Explanation:

The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.

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jek_recluse [69]

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erica [24]

Answer: The amount of $P_4O_{10}$ formed is 469.8 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol$

The given chemical reaction follows:

$4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)$

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of $P_4O_{10}$

So, 6.62 moles of phosphine will produce = $\frac{1}{4}\times 6.62=1.655mol$ of $P_4O_{10}$

Now, calculating the mass of $P_4O_{10}$ by using equation 1:

Molar mass of $P_4O_{10}$ = 283.9 g/mol

Moles of $P_4O_{10}$ = 1.655 moles

Putting values in equation 1, we get:

$1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g$

Hence, the amount of $P_4O_{10}$ formed is 469.8 grams.

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