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Veseljchak [2.6K]
3 years ago
5

Please help as soon as possible please and thank you.

Chemistry
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

The answer is A. I took the test today.

Explanation:

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Enter your answer in the provided box. The atomic masses of 63Cu (69.17 percent) and 65Cu (30.83 percent) are 62.94 and 64.93 am
dybincka [34]

<u>Answer:</u> The average atomic mass of copper is 63.55 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

  • <u>For _{29}^{63}\textrm{Cu} isotope:</u>

Mass of _{29}^{63}\textrm{Cu} isotope = 62.94 amu

Percentage abundance of _{29}^{63}\textrm{Cu} = 69.17 %

Fractional abundance of _{29}^{63}\textrm{Cu} isotope = 0.6917

  • <u>For _{29}^{65}\textrm{Cu} isotope:</u>

Mass of _{29}^{65}\textrm{Cu} isotope = 64.93 amu

Percentage abundance of _{29}^{65}\textrm{Cu} = 30.83 %

Fractional abundance of _{29}^{65}\textrm{Cu} isotope = 0.3083

Putting values in equation 1, we get:

\text{Average atomic mass of Copper}=[(62.94\times 0.6917)+(64.93\times 0.3083)]\\\\\text{Average atomic mass of copper}=63.55amu

Hence, the average atomic mass of copper is 63.55 amu.

4 0
3 years ago
Liquids:
swat32

Answer:

The correct answer is B

Explanation:

3 0
3 years ago
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Scientific notation.
stiv31 [10]

Answer:

1.85 x 10^ -5

Explanation:

8 0
3 years ago
Be sure to answer all parts. the equilibrium constant (kp) for the reaction below is 4.40 at 2000. k. h2(g) + co2(g) ⇌ h2o(g) +
svetoff [14.1K]
When we have the balanced reaction equation is:

H2(g)  + CO2(g) ↔ H2O(g)  + CO (g)

a) first, to calculate ΔG° for the reaction: 

we will use this formula:

ΔG° = -RT㏑Kp

when R is R- rydberg constant = 8.314J/mol.K

and T is the temperature in Kelvin = 2000 K

and Kp = 4.4 

so, by substitution:

ΔG° = - 8.314 *2000 *㏑4.4

        = - 24624 J/mol = - 24.6 KJ/mol

b) to calculate ΔG so, we will use this formula:

ΔG = ΔG° + RT㏑Qp

So we need first, to get Qp from the reaction equation:

when Qp = P products / P reactants

                 =  PH2O*PCO / PH2 * PCO2

                 = (0.66 atm * 1.2 atm) / (0.25 * 0.78)

                 = 4.1 

so by substitution:

ΔG = -24624 + 8.314*2000*㏑4.1

      = -1162 J/mol = - 1.16 KJ/mol
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4 years ago
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The sun earth and moon is close in?
ladessa [460]

Answer:

jupter

Explanation:

8 0
3 years ago
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