How can a sudden surge of fresh water from an estuary pollute the ocean?

diamong [38]

E. A direct measure of internal kinetic energy

7 0

3 years ago

A solution is prepared by pipetting 20.00 mL of a 1.00 x 10-3 M NaCl solution into a 250.0 mL volumetric flask. The flask is dil

oee [108]

Answer:

The concentration of the newly prepared solution is 8 * 10^-5 M

Explanation:

The most common way to solve this problem is to use the following formula

c1*V1 = c2*V2

with c1 = 1 * 10^-3 M

with V1 = 20 mL = 0.02 L

with c2 = TO BE DETERMINED

with V2 = 250 mL = O.250 L

10^-3 * 0.02 L = c2 * 0.250L

0.00002 = c2 * 0.250L

c2 = 0.00002 / 0.250 L

c2 = 0.00008 M = 8 * 10^-5 M

The concentration of the newly prepared solution is 8 * 10^-5 M

3 0

3 years ago

(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.

svp [43]

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

5 0

2 years ago

If you had a mixture of ethanol, glycerol, ethylene glycol, methanol, and water, how could you separate these five substances ou

leonid [27]

Fractional distillation

Explanation:

The best way to separate the mixtures out is through the process of fractional distillation.

In fractional distillation, liquid - liquid mixtures are separated based on the differences in boiling point of their components. Let us examine the boiling points of the component of the mixtures:

Ethanol 78⁰C

Glycerol 290⁰C

Ethylene glycol 197.6⁰C

Methanol 64.7⁰C

Water 100⁰C

We see that the liquids in the mixture have different boiling points. In this process, the mixture is heated in a distillation column. When the boiling point of any component is reached, it will rise up in the column and can be channeled to a condenser where it is cooled and collected.

The liquid with the least boiling point is first separated with the one with the highest boiling is recovered last:

Order of recovery;

Methanol 64.7⁰C

Ethanol 78⁰C

Water 100⁰C

Ethylene glycol 197.6⁰C

Glycerol 290⁰C

Learn more:

Physical properties brainly.com/question/10972073

#learnwithBrainly

3 0

3 years ago

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0

3 years ago