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strojnjashka [21]
2 years ago
9

What can you infer from the fact that nitrogen gas is useless to animals?

Chemistry
1 answer:
makvit [3.9K]2 years ago
3 0

Answer:

Denitrification changes nitrogen into a gaseous form.

Explanation:

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EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?
AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq).

The equilibrium constant for this reaction is called the self-ionization constant of water:

K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}].

Note that water isn't part of this constant.

The value of K_w at 25 °C is 10^{-14}. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}
  • However, [\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3} for pure water.
  • As a result, K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14} at 25 °C.

Back to this question. [\text{H}^{+}] is given. 25 °C implies that K_w = 10^{-14}. As a result,

\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}.

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A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC
diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 25^oC=273+25=298K

V = volume = 4.0 L

n_1 = moles of SO_2 = 0.20 mol

n_2 = moles of O_2 = 0.20 mol

Now put all the given values in the above expression, we get:

P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}

P=2.4atm

Thus, the pressure in the flask after reaction complete is, 2.4 atm

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