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yan [13]
2 years ago
5

The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what wi

ll the abundance be after 4.00 hours? To solve this problem, first use equation (7) to determine k for 64Cu; then use this k value in equation (6) to obtain the amount of 64Cu, A, remaining after 4.00 hours if the amount present at the start, A0, is 845 ppm.
Chemistry
1 answer:
Kisachek [45]2 years ago
7 0

Answer:

A = 679.2955 ppm

Explanation:

In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:

A = A₀ e^-kt

This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:

k = ln2 / t₁/₂

Replacing the given data we have:

k = ln2 / 12.7

k = 0.0546

Now, let's get the concentration of Cu:

A = 845 e^(-0.0546*4)

A = 845 e^(-0.2183)

A = 845 * 0.8039

A = 679.2955 ppm

This would be the concentration after 4 hours

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w/w percentage <span>
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70% HNO₃ contains by mass means every 100 g of sample has 70 g of HNO₃.</span><span>

The mass of solution = 103.8 g
Hence the mass of HNO₃ = 103.8 g x 70%</span><span>
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A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reacta
GalinKa [24]
Answer:

<span>23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. </span>

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula 

number of moles = mass in grams / molar mass


CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
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O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

3 0
2 years ago
6.An acid-base indicator is usually a weak acid with a characteristic color in the protonated and deprotonated forms. Because br
Eddi Din [679]

Answer:

At equal concentration of HBCG and BCG^-, the colour is green. This colour first appears at pH = 3.8

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HBCG is an indicator that is prepared by dissolving the solid in ethanol.

Since

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If pH = 3.8

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6 0
2 years ago
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