All categories

2 years ago

What is the pH of a solution that contains 0.83M HCN (Ka = 4.9x10-10) and 0.64 M potassium cyanide?

Chemistry

1 answer:

oksian1 [2.3K]2 years ago

5 0

Answer:

pH= 9.2

Explanation:

Henderson hasselbach equation

pKa= log Ka= log (4.9 x 10^-10)=9.3

$pH=Pka+log \frac{[A-]}{[HA]}$

$pH=9.3+log \frac{[CN-]}{[HCN]}$

$pH=9.3+log \frac{[0.64 M]}{[0.83 M]}$

pH= 9.2

You might be interested in

Use the Periodic Table. Choose the correct electron configuration of carbon using the data shown in the electron configuration c

AleksAgata [21]

I think I have done a question like this before and i'm pretty sure your answer would be the 2nd one 1s22s22p4. I'm not 100% sure but try it at least.

4 0

2 years ago

Read 2 more answers

Say I grab a bunch of aluminum foil (9.5g) and ball it up. Aluminum has a specific heat of .9J/g*C. How much would the aluminum’

Brrunno [24]

Answer:

ΔT = 0.78 °C

Explanation:

Given data:

Mass of Al = 9.5 g

Specific heat capacity of Al = 0.9 J/g.°C

Temperature change = ?

Heat added = 67 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

67 J = 9.5 g × 0.9 j/g.°C × ΔT

67 J = 85.5 j/°C × ΔT

ΔT = 67 J / 85.5 j/°C

ΔT = 0.78 °C

7 0

2 years ago

Joan needs a 0.053-M solution of HCl, but only has access to 2.12-M HCl, a 10-mL graduated pipet, and a 25-mL volumetric flask.

dmitriy555 [2]

Answer:

Joan should measure 0.625 mL of 2,12 M HCl solution to create her desired solution.

Explanation:

Molarity of HCl Joan has access to = $M_1=2.12 M$

Volume of 2.12 M of HCl Joan use = $V_1=?$

Molarity of HCl Joan desired = $M_2=0.053 M$

Volume of 0.053 M of HCl Joan can prepare = $V_2=25 mL$

$M_1V_1=M_2V_2$

$V_1=\frac{M_2V_2}{M_1}$

$=\frac{0.053 M\times 25 mL}{2.12 M}=0.625 mL$

Joan should measure 0.625 mL of 2,12 M HCl solution to create her desired solution.

6 0

3 years ago

A sample of ethanol has a volume of 7.5 mL and a mass of 5.85 g. A sample of benzene also has a volume of 7.5 ml,

AysviL [449]

Answer:

Benzene

Explanation:

You need to calculate the densities for each compound.

EtOH = 5.85/7.5 = 0.78 g/mL

Benzene = 6.60/7.5 = 0.88 g/mL

0.88 > 0.78, thus benzene is denser than ethanol.

3 0

2 years ago

Read 2 more answers

Suppose protons were emitted rather than electrons. How would this affect the experiment?

Vera_Pavlovna [14]

Suppose protons were emitted rather than electrons then it affects the experiment as the mean velocity of proton will be less then mean velocity electron .

The mass of proton is greater than the mass of electrons but charge of electron is equal to the charge of proton . So , due to difference in the mass of electron and proton there will be some effects. We can conserve the electric energy which is equal to the qe .

The kinetic energy = 1/2 $mv^{2}$

Now changing the electric potential energy into kinetic energy

v = √2qe/m

The mean velocity of proton will be less then mean velocity electron .

To learn more about proton

brainly.com/question/1252435

#SPJ4

6 0

1 year ago