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3 years ago

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

1 answer:

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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3 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

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Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

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2 years ago

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3 years ago

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Answer:

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See attached file

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3 years ago