All categories

2 years ago

What is the molarity of solution prepared by dissolving 11.75 g of KNO3 in enough water to produce 2.00 L of solution?

1 answer:

3 0

The molar mass of the compound potassium nitrate, KNO3 is equal to 101.1032 g/mol. Then, we determine the number of moles present in the given amount,
n = 11.75g / (101.1032 g/mol) = 0.116 mol
Then, molarity is calculated by dividing the number of moles by the volume of the solution. The answer is therefore 0.058 M.

You might be interested in

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

jolli1 [7]

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

Where:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

7 0

2 years ago

g Select the irreversible reactions of glycolysis. conversion of glucose to glucose 6‑phosphate by hexokinase conversion of gluc

alexdok [17]

Answer:

1) Conversion of glucose to glucose 6-phosphate by hexokinase

2) Conversion of fructose 6-phosphate to fructose 1,6-biphosphate by phosphofructokinase

3) Conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase

Explanation:

There are 10 steps in the glycolysis pathway, three of which are irreversible. The enzymes controlling these reactions have not only catalytic properties but the irreversibility of the reaction gives them regulatory properties as well. These reactions serve as control points in the pathway.

3 0

2 years ago

An atom of lithium-7 has an equal number of

slavikrds [6]

The answer is (2). The Lithium has 3 protons and 4 neutrons. For every neutral atom, it will have the same number of protons and electrons. Because the proton has one positive charge and electron has one negative charge and neutron does not have charge.

8 0

2 years ago

When a pH probe is inserted into a solution containing the chloride ion it is neutral. What is the pH of a solution containing t

vivado [14]

Answer:

a. the solution will be weakly basic.

b. Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.

Explanation:

a. The fluoride ion (F⁻) reacts with water thus:

F⁻ + H₂O → HF + OH⁻

That means that fluoride ions produce OH⁻ ions in solution doing the solution will be weakly basic.

b. The acidic equilibrium of NH₄⁺ is:

NH₄⁺ ⇄ NH₃ + H⁺ with a ka of 5,6x10⁻¹⁰.

The basic equilibrium of CN⁻ is:

CN⁻ + H₂O → HCN + OH⁻ with a kb of 2x10⁻⁵

That means that the production of OH⁻ from CN⁻ is higher than production of H⁺ from NH₄⁺. The CN⁻ is a stronger base than NH₄⁺ is an acid.

Thus, the pH of a salt solution of NH₄CN would be Greater than 7 because CN⁻ is a stronger base than NH₄⁺ is an acid.

I hope ot helps!

3 0

2 years ago

If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

2 years ago