When a solute molecule is solvated, is energy released or absorbed?

Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such

8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

1 A = 1 C/s
The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

$19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g$

5 0

2 years ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

2 years ago

1. Which liquid sample is a pure substance?

IRISSAK [1]

The whole Activity , poem and paragraph is missing in the question.

Answer:

(1) Liquid A

(2) Solid A

Explanation:

Using this part of the given poem

Substances and mixtures behave differently,

During boiling and melting most especially

Boiling point of substance is fixed while mixture is not

Substance melts completely but mixture does not

The boiling point of the Pure substance remain fixed after reaching its boiling point this is shown by Liquid A

Solid A is melting completely so Solid A is a pure substance.

6 0

2 years ago

Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

Ghella [55]

Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

5 0

2 years ago

Read 2 more answers

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago