All categories

3 years ago

The weak ionization constant (Ka) for HCN is equal to:

Chemistry

1 answer:

denis-greek [22]3 years ago

3 0

Answer:

Ka = [H⁺] × [CN⁻] / [HCN]

Explanation:

Cyanhydric acid is a weak acid, according to the following equation:

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

The acid ionization constant (Ka) is equal to the product of the concentrations of the ions raised to the stoichiometric coefficients divided by the concentration of the undissociated acid raised to its stoichiometric coefficient.

Ka for HCN is:

Ka = [H⁺] × [CN⁻] / [HCN]

You might be interested in

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

In poor quality fireworks what will you notice and why

anastassius [24]

The best answer I could find was when you Google it, that the fuse is of poor quality. I cannot leave you a link, but you can find it for yourself. Put in poor quality fireworks and all sorts of things will pop up. No pun intended.

8 0

3 years ago

Read 2 more answers

What is Double Decomposition Reaction? _ _ _ koi Zinda bacha hai?

Sholpan [36]

Answer:

It’s when two different elements in a compound get switched

Explanation:

4 0

2 years ago

Explain the connection between periods and energy levels.

Rainbow [258]

Answer:

The atoms in the first period have electrons in 1 energy level. The atoms in the second period have electrons in 2 energy levels. The atoms in the third period have electrons in 3 energy levels. The atoms in the fourth period have electrons in 4 energy levels.

7 0

3 years ago

The AE of a system that releases 12.4 J of heat and does 4.2 J of work on its surroundings It is_______ jA. 16.6 B. 12.4 C. 4.2

andre [41]

ANSWER

EXPLANATION

Given that

The energy released by the system is 12.4J

Work done on the surrounding is 4.2J

Follow the steps below to find the change in energy

In the given data, energy is said to be released to the surroundings

Recall, that exothermic reaction is a type of reaction in which heat is released to the surroundings. Hence, change in enthalpy is negative

Step 1; Write the formula for calculating change in energy

$\Delta E\text{ }=\text{ q }+\text{ w}$

Since heat is released to the surrounding, then q = -12J

Recall, that work done by the system on the surroundings is always negative

Hence, w = -4.2J

Step 2; Substitute the given data into the formula in step 1

$\begin{gathered} \text{ }\Delta E\text{ = q + w} \\ \text{ }\Delta E\text{ }=\text{ -12.4 }+\text{ \lparen-4.2\rparen} \\ \text{ }\Delta E\text{ = -12.4 - 4.2} \\ \text{ }\Delta E\text{ }=\text{ -16.6J} \end{gathered}$

Therefore, the change i

3 0

1 year ago