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myrzilka [38]
3 years ago
14

Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction.

Chemistry
1 answer:
krok68 [10]3 years ago
5 0
You didn't include much information as to what you were referring too, however I answered a similar question recently to a friend so I'm hoping this is the same question in complete as to yours, if not I apologize.

a.) 2Na(s) + Cl2(g) -----> 2NaCl(s) 
Na --> Na+1 is oxidized 
Cl2 --> Cl- is reduced 

b.) 2HNO3(aq) + 6HI(aq) ----> 2NO(g) + 3I2(s) + 4H2O(l) 
N+5 --> N+2 is reduced 
I-1 --> I2 is oxidized 
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8 0
3 years ago
Read 2 more answers
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
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victus00 [196]

Answer:

Got what wrong? A question?

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