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mixer [17]
2 years ago
10

The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

reviation (w/v) is also common. How many grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution
Chemistry
1 answer:
Talja [164]2 years ago
6 0

The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abbreviation (w/v) is also common. 262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution

<h3>Define Solute</h3>

A solute is a material that dissolves in a solution. The amount of solvent present in fluid solutions is greater than the amount of solute. The two most common examples of solutions in daily life are salt and water. Salt is the solute because it dissolves in water.

<h3>forms of ratios for product concentration or yield:-</h3>
  • w/v:- Weight by volume or weight per volume are the terms used. Any solid compound's concentration in a liquid can be calculated using it. It is measurable in gm/ml.
  • Weight by weight ratio is referred to as w/w.It is employed to determine the final yield of the compound obtained from the starting compound. as in —mg/—gm.

       It provides the real yield of the substance or item.

  • Volume/volume. It is used to specify a liquid's composition or percent in a liquid compound.

using w/v we can calculate the weight of sucrose:-

40.0% means 40 g sucrose/ 100 g solution

40.0g sucrose x (655/100)=grams of sucrose

262  grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution.

Learn more about Solute here:-

brainly.com/question/14397121

#SPJ4

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maksim [4K]

Answer:

6.022  ×10(index 23) / 7.5 = 0.8293 ×10(index 23)

Explanation:

molar mass of C =  12gmol

therefore in 12g of C there is one mole or an amount of 6.022 ×10(index 23)

∴12g/6.02210(index 23) ×1.6g

3 0
3 years ago
Determine the molar concentration of na+ and po4 3- in a 2.25 M Na3 PO4 solution
muminat

Answer:

A. The concentration of Na^+ in the solution is 6.75 M.

B. The concentration of PO4^3- in the solution is 2.25 M.

Explanation:

We'll begin by writing the balanced dissociation equation for Na3PO4.

This is illustrated below:

Na3PO4 will dissociate in solution as follow:

Na3PO4(aq) —> 3Na^+(aq) + PO4^3-(aq)

Thus, from the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+ and 1 mole of PO4^3-

A. Determination of the concentration of Na+ in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+.

Therefore, 2.25 M Na3PO4 solution will produce = (2.25 x 3) /1 = 6.75 M Na^+.

Therefore, the concentration of Na^+ in the solution is 6.75 M

B. Determination of the concentration of PO4^3- in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 1 mole of PO4^3-

Therefore, 2.25 M Na3PO4 solution will also produce 2.25 M PO4^3-.

Therefore, the concentration of PO4^3- in the solution is 2.25 M.

7 0
3 years ago
11. The modern periodic table is arranged in order of increasing atomic
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Answer:

number. so the option is c

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Explanation:

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Chemistry, please help! Thanks.
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Answer:

d. The gold(III) ion is most easily reduced.

Explanation:

The standard reduction potentials are

Au³⁺ + 3e⁻ ⟶ Au;   1.50 V

Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V

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