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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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What is the diffrence between a mixture and a compound
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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

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Pani-rosa [81]

Answer:

0.37atm

Explanation:

Given parameters:

Initial pressure  = 0.25atm

Initial temperature  = 0°C  = 273K

Final temperature  = 125°C  = 125 + 273  = 398K

Unknown:

Final pressure  = ?

Solution:

To solve this problem, we use a derivative of the combined gas law;

           \frac{P1}{T1}  = \frac{P2}{T2}

  P and T are pressure and temperature

  1 and 2 are initial and final values

        \frac{0.25}{273}   = \frac{P2}{398}  

         P2  = 0.37atm

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