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erastovalidia [21]
3 years ago
8

The atomic weight of copper is 63.546 amu. The two naturally occuring isotopes of copper have the following masses 63Cu, 62.9298

amu, 65Cu, 64.9278 amu. Calculate percent of 63Cu in naturally occuring copper​
Chemistry
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

The percent isotopic abundance of Cu-65 is 30.79 %.

The percent isotopic abundance of Cu-63 is 69.21 %.

Explanation:

we know there are two naturally occurring isotopes of carbon, Cu-63 (62.9298 amu)  and Cu-65 (64.9278 amu)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 62.9298 amu

1-x for isotopes having mass 64.9278 amu

The average atomic mass of copper is 63.546 amu.

we will use the following equation,

62.9298 x + 64.9278 (1-x) = 63.546

62.9298x +  64.9278 -  64.9278x = 63.546

62.9298x - 64.9278x = 63.546  - 64.9278

-1.998 x = -1.3828

x=-1.3828 / -1.998

x= 0.6921

0.6921 × 100 = 69.21 %

69.21 % is abundance of Cu-63 because we solve the fraction x.

now we will calculate the abundance of Cu-65.

(1-x)

1-0.6921 = 0.3079

0.3079 × 100 = 30.79 %

30.79 % for Cu-65.

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3 years ago
In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera
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Answer : The enthalpy change of dissolution of KClO_4 is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 1.55J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 107.70 g

\Delta T = change in temperature = T_2-T_1=(22.80-20.34)=2.46^oC

Now put all the given values in the above formula, we get:

q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]

q=1112.3J=1.1123kJ

Now we have to calculate the enthalpy change of dissolution of KClO_4

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of KClO_4 = 2.84 g

Molar mass of KClO_4 = 138.55 g/mol

\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole

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Therefore, the enthalpy change of dissolution of KClO_4 is 54.3 kJ/mole

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If an object is only partially submerged in a fluid, which of the following is true?
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Answer:

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We are given an unknown sample with a mass of 23.5 g and a volume of 2.00 mL and we want to determine its density.

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