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AURORKA [14]
4 years ago
5

What is the percent composition of CuCl2?

Chemistry
2 answers:
-Dominant- [34]4 years ago
8 0
CuCl_2
There is 2 times more Cl than Cu.

So for 1 :
Cu : \frac{1}{3}
Cl : \frac{2}{3}

So percent composition :
Cl ⇒ 66%
Cu ⇒ 33%
allochka39001 [22]4 years ago
4 0
CuCl2 is Dichloride Copper, or Copper II Chloride. 
In this Molecule, there is 1 atom of Copper for every 2 atoms of Chlorine.
If you go to the periodic table, you'll see that copper has a mass of 63.5 amu, and the chlorine 35.45 amu.
Which mean that the mass of CuCl2 is 63.5 + 35.45*2 = 63.5 + 70.9 = 134.4 g/mol.
Then, you find the ration of the mass of each atom and multiply by 100.
Ratio of Copper: \frac{63.5}{134.4} * 100 and you get about 47.2% 
Ration of chlorice (Cl2): \frac{70.9}{134.4} * 100 and you get about 52.8%.

So, the percent composition of CuCl2 is 47.2% of Copper and 52.8% of chlorine.

Hope this Helps :)
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Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

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In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

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3 years ago
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nirvana33 [79]

Answer:

44.8 L

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T = temperature (K)

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7 0
3 years ago
For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula
bekas [8.4K]

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CO(g) + 2H₂(g) ⇄ CH₃OH(g)

 

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²

</span>[H₂<span>]=0.072 M
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[CH</span>₃OH]= 0.030 M

 

From substitution,

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<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².

<span> </span>

3 0
3 years ago
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never [62]
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