Ask question

Ask question

All categories

borishaifa [10]

3 years ago

14

What is the pH of a 9.50×10−2 M ammonia solution??

1 answer:

Darina [25.2K]3 years ago

4 0

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

c=9,50*10⁻² mol/L

Ammonia solution is a weak electrolyte.
The dissociation constant of ammonium hydroxide is:
K=1.78*10⁻⁵

[OH⁻]=√(Kc)

pH=14-pOH=14+lg[OH]⁻
pH=14+lg(√(Kc))

pH=14+lg(√(1.78*10⁻⁵*9.50*10⁻²))=8.23
pH=8.23

You might be interested in

grandma baked cookies. she gave 7 to sandy. she gave half of what were left, plus one to sue. she gave 3 to stan. then she gave

scoray [572]

I believe your answer is 23.
Credit: answers.yahoo.com

Hope this helps!

8 0

3 years ago

Which is NOT a way nitrogen can be fixed or made more useful?

stich3 [128]

This doesn’t have a multiple choice part to answer.

7 0

3 years ago

The mass and volume of each box is given in the chart below. Calculate the density of each box. Density = Mass/ volume which box

stira [4]

Box C will have the greatest density.

All boxes have the same volume.

Explanation:

We calculate the density using the following formula:

density = mass / volume

density of Box A = 10 g / 20 cm³ = 0.5 g/cm³

density of Box B = 30 g / 20 cm³ = 1.5 g/cm³

density of Box C = 170 g / 20 cm³ = 8.5 g/cm³

Box C will have the greatest density.

All boxes have the same volume.

Learn more about:

density

brainly.com/question/952755

#learnwithBrainly

7 0

3 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

<u>Answer:</u> The entropy change of the process is $2.0\times 10^2J/K$

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

4 0

3 years ago

List the following elements in order from smallest atomic radii to largest:<br> Cr, Fe, Zn, Sc

Ber [7]

Sc(21), Cr(24), Fe(26), Zn(30)

4 0

3 years ago