Answer:
pOH = 11.5
[H⁺] = 0.003 M
[OH⁻] = 3 × 10⁻¹² M
Explanation:
The computation is shown below:
Given that
pH = 2.5
Based on the above information
We know that
pH + pOH = 14 ⇒ pOH = 14 - pH
pOH = 14 - 2.5
pOH = 11.5
[H⁺] = 10^(-pH) = 10^(-2.5)
[H⁺] = 0.003 M
[OH⁻] = 10^(-pOH)
= 10^(-11.5)
= 3 × 10⁻¹² M
[OH⁻] = 3 × 10⁻¹² M
Hence, the above represents the answer
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.
∴
(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
It contributes by breathing. CO2 is carbon dioxide we breath it out and plants convert it into oxygen.
