What Does The Atomic Radius Change Down A Column Of The Periodic Table

Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit

rosijanka [135]

Explanation:

The relation between $K_a\&K_b$ is given by :

$K_w=K_a\times K_b$

Where :

$K_w=1\times 10^{-14}$ = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = $K_{a1}=1.4\times 10^{-2}$

The value of $K_{b1}$ :

$1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}$

The value of the second ionization constant of sodium sulfite = $K_{a2}=6.3\times 10^{-8}$

The value of $K_{b2}$ :

$1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}$

3 0

3 years ago

You have a 28.2-g sample of a metal heated to 95.2°c. you drop it in a calorimeter with 100. g of water at 25.1°c. the final tem

Vlad [161]

The heat lost by the metal should be equal to the heat gained by the water. We know that the heat capacity of water is simply 4.186 J / g °C. Therefore:

100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp * (95.2°C - 31°C)

4 0

3 years ago

Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre

weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

$\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)$

which is equivalent to

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

The question states that the second equation has an enthalpy, or "heat", of neutralization of $-56.2 \; \text{kJ}$ . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce $56.2 \; \text{kJ}$ or $56.2 \times 10^{3}\; \text{J}$ of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release $0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J}$ of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or $1.00 \times 10^{3} \; \text{ml}$ of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of $1.00 \times 10^{3} \; \text{g}$ .

The solution has a specific heat of $4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . The solution thus have a heat capacity of $4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}$ . Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of $4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}$ , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. $1.405 \times 10^{4} \; \text{J}$ are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.