A compound has an empirical formula of

Chemistry

2 answers:

Ksju [112]3 years ago

8 0

Answer: C12H12N2O2 (Cyclomethyltryptophan)

chubhunter [2.5K]3 years ago

3 0

Answer:

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,

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

8 0

3 years ago

A solution is prepared by adding 0.0231moles of H3O+ ions to 3.33L of water. What is the pH of this solution

MrMuchimi

Answer:

2.15

Explanation:

For this question, we have to remember the pH formula:

$pH~=~-Log[H_3O^+]$

By definition, the pH value is calculated when we do the -Log of the concentration of the hydronium ions ( $H_3O^+$ ). So, the next step is the calculation of the concentration of the hydronium ions. For this, we have to use the molarity formula: