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Andre45 [30]
3 years ago
10

Which of the following statements, if true, would support the claim that the NO3− ion, represented above, has three resonance st

ructures? The NO3− ion is not a polar species. A The oxygen-to-nitrogen-to-oxygen bond angles are 90°. B One of the bonds in NO3− is longer than the other two. C One of the bonds in NO3− is shorter than the other two.
Chemistry
1 answer:
tester [92]3 years ago
7 0

Answer:

One of the bonds in nitrate is shorter than the other two.

Explanation:

We would firstly need to draw the Lewis structure for nitrate anion. To do this, let's follow the standard steps:

  • calculate the total number of valence electrons: five from nitrogen, each oxygen contributes 6, so a total of 18 from oxygen atoms, as well as one from the negative charge, we have a total of 24 valence electrons;
  • assign the central atom, usually this is the atom which is single; in this case, we have nitrogen as our central atom;
  • assign single bonds to all the terminal atoms (oxygen atoms);
  • assign octets to the terminal atoms and calculate the number of electrons assigned;
  • the number of electrons assigned is 24, so no lone pairs are present on nitrogen;
  • calculate the formal charges: each oxygen has a formal charge of -1 (formal charge is calculated subtracting the sum of lone pair electrons and bonds from the number of valence electrons of that atom); nitrogen has a formal charge of +2;
  • nitrogen doesn't have an octet as well, so we'll both minimize its formal charge and make it obtain an octet if we make one double bond N=O.

Therefore, we may have 3 resonance structures, as this double bond might be formed with any of the 3 oxygen atoms.

By definition, double bonds are shorter than single ones, so one of the bonds is shorter than the other two.

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its a

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A mixture of gases is analyzed and found to have the following composition in mol %: CO2 12.0 CO 6.0 CH4 27.3 H2 9.9 N2 44.8 a)
vichka [17]

Answer:

a) CO₂: <em>21,9%; </em>CO: <em>7,0%; </em>CH₄: <em>18,2%; </em>H₂: <em>0,8%; </em>N₂: <em>52,1%</em>

b) 24,09 g/mol

Explanation:

a) It is posible to obtain the composition of the gas mixture in weight% using molecular mass of each compound, thus:

12% CO₂×\frac{44,01g}{1mol} = <em>528,1 g</em>

6% CO×\frac{28,01g}{1mol} = <em>168,1 g</em>

27,3% CH₄×\frac{16,05g}{1mol} = <em>438,2 g</em>

9,9% H₂×\frac{2,02g}{1mol} = <em>20,0 g</em>

44,8% N₂×\frac{28g}{1mol} = <em>1254,4 g</em>

The total mass of the gas mixture is:

528,1g + 168,1g + 438,2g + 20,0g + 1254,4g = <em>2408,8 g</em>

Thus composition of the gas mixture in weight% is:

CO₂: \frac{528,1g}{2408,8g}×100 = <em>21,9%</em>

CO: \frac{168,1g}{2408,8g}×100 = <em>7,0%</em>

CH₄: \frac{438,2g}{2408,8g}×100 = <em>18,2%</em>

H₂: \frac{20,0g}{2408,8g}×100 = <em>0,8%</em>

N₂: \frac{1254,4g}{2408,8g}×100 = <em>52,1%</em>

b) The average molecular weight of the gas mixture is determined with mole % composition, thus:

0,12×44,01g/mol + 0,06×28,01g/mol + 0,273×16,05g/mol + 0,099×2,02g/mol + 0,448×28g/mol = <em>24,09 g/mol</em>

I hope it helps!

6 0
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