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Vadim26 [7]
3 years ago
10

Please help me with this chemistry question, image attached

Chemistry
1 answer:
ss7ja [257]3 years ago
6 0

average atomic mass of Cu = 63.54

Explanation:

average atomic mass =  atomic mass of isotope 1 × abundance of isotope 1 + atomic mass of isotope 2 × abundance of isotope 2 +...+ atomic mass of isotope n × abundance of isotope n

average atomic mass of Cu = atomic mass of Cu-63 × abundance of isotope Cu-63 + atomic mass of isotope Cu-65 × abundance of isotope Cu-65

average atomic mass of Cu = 62.93 × (69.17 / 100) + 64.93 × (30.83 / 100)

average atomic mass of Cu = 63.54

Learn more about:

average atomic mass

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What type of molecular model shows all the atoms and bonds in an organic molecule
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Answer:

Structural formula shows the atoms and bondsin an organic compound.

Explanation:

Structural formula of methane shows

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3 years ago
The most common source of copper (Cu) is the mineral chalcopyrite (CuFeS2). How many kilograms of chalcopyrite must be mined to
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3 years ago
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Question 26 Suppose a flask is filled with of and of . The following reaction becomes possible: The equilibrium constant for thi
Blizzard [7]

Answer:

0.36 M

Explanation:

There is some info missing. I think this is the complete question.

<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible: </em>

<em>N₂(g) +O₂(g) ⇄ 2 NO(g) </em>

<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask.  Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>

<em />

Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:

[N₂] = 0.30 mol / 0.250 L = 1.2 M

[NO] = 0.70 mol / 0.250 L = 2.8 M

We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.

    N₂(g) +O₂(g) ⇄ 2 NO(g)

I    1.2        0              2.8

C  +x         +x            -2x

E  1.2+x      x           2.8 - 2x

The equilibrium constant (K) is:

K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}

Solving for x, the positive one is x = 0.3601 M

[O₂] = 0.3601 M ≈ 0.36 M

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