Which two properties are characterized of ionic compounds

A student prepares a aqueous solution of trimethylacetic acid . Calculate the fraction of trimethylacetic acid that is in the di

irinina [24]

The question is incomplete, here is the complete question:

A student prepares a 0.21 mM aqueous solution of trimethylacetic acid. Calculate the fraction of trimethylacetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 3 significant digits.

Answer: The percent of dissociation of trimethylacetic acid is 19.5 %

Explanation:

We are given:

Concentration of trimethylacetic acid = 0.21 mM = 0.00021 M (Conversion factor: 1 mole = 1000 millimoles )

The chemical equation for the dissociation of trimethylacetic acid follows:

$C_4H_9COOH\rightleftharpoons C_4H_9COO^-+H^+$

Initial: 0.00021

At eqllm: 0.00021-x x x

The expression of $K_a$ for above equation follows:

$K_a=\frac{[C_4H_9COO^-][H^+]}{[C_4H_9COOH]}$

We know that:

$K_a\text{ for }C_4H_9COOH=1\times 10^{-5}$

Putting values in above expression, we get:

$1\times 10^{-5}=\frac{x\times x}{(0.00021-x))}\\\\x=0.000041,-0.000051$

Neglecting the negative value of 'x' because concentration cannot be negative.

To calculate the fraction of dissociation, we use the equation:

$\text{Percent of dissociation}=\frac{[H^+]}{[C_4H_9COOH]}\times 100$

Putting values in above equation, we get:

$\text{Percent of dissociation}=\frac{0.000041}{0.00021}\times 100\\\\\text{Percent of dissociation of trimethylacetic acid}=19.5%$

Hence, the percent of dissociation of trimethylacetic acid is 19.5 %

8 0

3 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

3 years ago

For the reaction 2A(g) â B(g), the equilibrium constant is Kp = 0.76. A reaction mixture initially contains 4.0 atm of gas (PA =

timofeeve [1]

Answer: (a) The reaction mixture will proceed toward products.

Explanation:

Equilibrium constant is defined as the ratio of pressure of products to the pressure of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$2A(g)\rightleftharpoons B(g)$

The expression for $Q_p$ is written as:

$Q_p=\frac{p_B}{(p_A)^2}$

$Q_p=\frac{(2.0)}{(2.0)^2}$

$Q_p=0.5$

$K_p=0.76$

Thus as $K_p>Q_p$ , the reaction will shift towards the right i.e. towards the product side.

7 0

3 years ago

An acetic acid buffer containing 0.50 M acetic acid (CH3COOH) and 0.50 M sodium acetate (CH3COONa) has a pH of 4.74. What will t

ad-work [718]

Answer:

pH = 4.71

Explanation:

We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:

pH = pKa + log [CH3COONa] / [CH3COOH]

Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.

As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:

pH = pKa + log 1

4.74 = pKa

To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl

Moles CH3COONa: Initial CH3COONa - Moles HCl.

Moles CH3COOH:

0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =

0.052 moles CH3COOH

Moles CH3COONa:

0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =

0.048 moles CH3COONa

Using H-H equation:

pH = 4.74 + log [0.048 moles] / [0.052 moles]

5 0

3 years ago

El plomo y el yodo forman dos compuestos. En uno, el porcentaje de en masa de plomo es del 44,94% y en el otro es de 62,02%. Cal

nikdorinn [45]

Answer:

0.8162 gramos de plomo por gramo de yodo

1.633 gramos de plomo por gramo de yodo

Explanation:

Asumiendo una base de 100 gramos para cada compuesto:

Primer compuesto:

Gramos plomo: 44.94g

Gramos de yodo: 100-44.94g = 55.06g

Así, la masa de plomo por gramos de yodo para el primer compuesto es:

44.94g plomo / 55.06g Yodo =

0.8162 gramos de plomo por gramo de yodo

Segundo compuesto:

Gramos plomo: 62.02g

Gramos de yodo: 100-62.02g = 37.98g

La masa de plomo por gramos de yodo para el segundo compuesto es:

62.02g plomo / 37.98g Yodo =

1.633 gramos de plomo por gramo de yodo

8 0

3 years ago