Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Yes.
A covalent bond holds it together, which is chemical.
Answer:
D. ![K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5B%5Ctext%7BNO%7D_%7B2%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BHNO%7D_%7B2%7D%5D%7D)
Explanation:
The general form of an equilibrium constant expression is
![K = \frac{[\text{Products}]}{[\text{Reactants}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5B%5Ctext%7BProducts%7D%5D%7D%7B%5B%5Ctext%7BReactants%7D%5D%7D)
In the equilibrium
HNO₂ ⇌ H⁺ + NO₂⁻
The products are H⁺ and NO₂⁻, and the reactant is HNO₂.
∴ ![K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5B%5Ctext%7BNO%7D_%7B2%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BHNO%7D_%7B2%7D%5D%7D)