All categories

3 years ago

What’s the answer? And if you find the answer in a pdf, comment the link.

Chemistry

1 answer:

dimulka [17.4K]3 years ago

4 0

You can simply look at the order of the elements on a periodic table. Therefore, it is B)

You might be interested in

What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe

Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ C CH3COOH + C CH3COONa = [ CH3COOH ] + [ CH3COO- ]

∴ C CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ C CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒ [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = C CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0

3 years ago

How does the appearance of the protist differ from that of the onion sample?

lana [24]

The appearance of the protist differ from that of the onion sample due to the presence of motile structures.

<h3>What is a Protist?</h3>

This is an eukaryote which could be unicellular or multicellular and examples include protozoa etc.

The major difference between a protist and onion sample is that protists have motile structures such as flagella, cilia etc while plant cells such as onions don't have.

Read more about Protist here brainly.com/question/2169979

#SPJ1

8 0

2 years ago

HURRY I DONT HAVE TIME Arrange the waves from highest energy to lowest energy.

ioda

Cab

Hope this helps :D

4 0

3 years ago

You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli

maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

$M_1V_1=M_2V_2$

where, $M_1$ is initial molarity and $M_2$ is the molarity after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.