Water samples containing dissolved substances are called _____ solutions. A. solvent B. aqueous C. anhydrous D. elementary

Which option names two processes that must occur to form soil? weathering breaks rocks into minerals and plants die and decay er

jeka57 [31]

Answer:

"weathering breaks rocks into minerals and plants die and decay" must occur to form soil

Explanation:

For the formation of the soil two process are mandatory and they are erosion as well as weathering. The weathering takes place near the surface of the soil, which can occur due to various physical and chemical processes. Being an important part of soil development. It may occur rapidly or can be a slow process for decades. Water and air helps in weathering. The transport is generally by wind or with the help of flow of water, ice, etc .,

8 0

3 years ago

Read 2 more answers

Help on this question please?

il63 [147K]

Answer:

Your answer would be A. The fireworks

Explanation:

The reason behind this is because fireworks do produce light and sound, but they weren't made to. They were made with no noise but now they are more spectacular and "Boom" in the air with beautiful light with colors.

I hope this helps you! If I am totally screwed up and wrong. please let me know in the comment section below! Have a great day, my friend!

8 0

3 years ago

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following v

marissa [1.9K]

Answer:

a) 2.79

b) 4.75

c) 6.59

d) 8.81

e) 11.03

f) 12.42

Explanation:

Step 1: Data given

Volume of acetic acid = 35.0 mL

Molarity of acetic acid = 0.150 M

Molarity of NaOH = 0.150 M

Step 2: pH after adding 0 mL of NaOH

HA + H2O ⇆ A- + H3O+

Initial concentration

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at the equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

We can write it like this

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

Step 3: pH after adding 17.5 mL of NaOH

This is the point of half-neutralization.

pH = pKa = -log (1.76*10^-5- 4.75

Step 4: pH after adding 34.5 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.0345 L

Moles NaOH = 0.005175 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.005175 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.005175 = 0.000075 moles

[OH-] = 0.005175 - 0.005175 = 0 moles

[A-] = 0.005175 moles

pH = pKa + log([A-]/[HA]

pH = 4.75 +log(0.005175/0.000075)

pH = 4.75 + 1.84 = 6.59

Step 4: pH after adding 35.0 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.035 L

Moles NaOH = 0.00525 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.00525 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.00525 = 0 moles

[OH-] = 0.00525 - 0.00525 = 0 moles

[A-] = 0.00525 moles

We have a solution of 0.00525 mol of A - in 70.0 mL

[A-] = 0.00525 moles / 0.070 L = 0.075 M

A- + H20 ⇆ HA + OH-

Initial moles A- = 0.0750 moles

initial moles HA = 0 moles

Initial moles OH- = 0 moles

Moles A- at the equilibrium = 0.0750 - x

Moles HA at the equilibrium = x

Moles OH- at the equilibrium = x

Kb = Kw/Ka = 10^-14 / 1.76*10^-5

Kb = 5.68 *10^-10

Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10

Since 0.0750 >>> x we can write

5.68*10^-10 = x²/0.0750

x = 6.53 *10^-6

[OH-] = x = 6.53 * 10^-6 M

pOH = - log (6.53*10^-6) = 5.19

pH = 14 - pOH = 14 - 5.19 = 8.81

Step 5: pH after adding 35.5 mL of NaOH

We add an excess moles of NaOH

We have an excess of 0.5 mL NaOH

Moles excess = 0.0005 L *0.150 M = 0.000075 moles

[OH-] = 0.000075 moles / (0.035+0.0355L)

[OH-] = 1.06 * 10^-3 M

pOH = -log( 1.06 *10^-3)

pOH = 2.97

pH = 14 -2.97 = 11.03

Step 6: pH after adding 50.0 mL of NaOH

We have an excess of 15.0 mL NaOH

Moles excess = 0.015 L *0.150 M = 0.00225 moles

[OH-] = 0.00225 moles / (0.035+0.050L)

[OH-] = 0.0265 M

pOH = -log( 0.0265M)

pOH = 1.58

pH = 14 -1.58 = 12.42

4 0

3 years ago

Calculate the molarity of 1 L of water at 20 °C given a density of 0.9982 g/mL.

MaRussiya [10]

Answer:

I have the same thing

Explanation:

7 0

3 years ago

1. An object takes 5 seconds to travel 25 meters. What is its speed?

Keith_Richards [23]

Answer:

s = 5m/s

Explanation:

Given data:

Time taken by object = 5 sec

distance cover by object = 25 m

Speed of object = ?

Solution:

Formula:

s = d/t

s = speed

d = distance

t = time

s = 25 m/ 5 s

s = 5m/s

8 0

3 years ago