Potential energy is what

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago

What is the number assigned to an element on the Periodic Table?

spayn [35]

Answer:

Element Atomic Number One number you will find on all periodic tables is the atomic number for each element.

Explanation:

hope it helps :))

6 0

3 years ago

Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0

3 years ago

What do Potassium and Krypton have in common and what are their differences based on their location on the periodic table?

Kamila [148]

Potassium (K) and Krypton (Kr) are in the same period, so they have the same number of electron shells. Potassium is more reactive than Krypton because it only has one valence electron.

K and Kr are both in Period 4, so they each have the n = 1, 2, 3, and 4 shells. Kr is less reactive than K because it already has a complete octet.

“Potassium (K) and Krypton (Kr) are in the same period, so they have the same number of electron shells. Krypton is more reactive than Potassium because it has a full outer shell” is incorrect because having a full outer shell means that the element is unreactive.

“Potassium (K) and Krypton (Kr) are in the same group, so they have the same number of valence electrons. They have different numbers of electron shells” is incorrect for three reasons:

1. K is in Group 1 and Kr is in Group 18.

2. K has one valence electron, and Kr has eight valence electrons.

3. They have the same number of electron shells.

“Potassium (K) and Krypton (Kr) are in the same group, so they have the same number of valence electrons. They have different numbers of electron shells” is incorrect for two reasons:

1. K and Kr have different numbers of valence electrons.

2. They have the same number of electron shells.

8 0

3 years ago

Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to

hram777 [196]

Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

$MCl_n=MCl_2$