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3 years ago

A potassium bromide solution is 8.30 % potassium bromide by mass and its density is 1.03 g/mL.

Chemistry

1 answer:

s344n2d4d5 [400]3 years ago

5 0

Answer:

3.52 g.

Explanation:

The density of potassium bromide solution is 1.03 g/mL, which means that the mass of 1.0 mL of the solution is 1.03 g.

Using cross multiplication:

1.0 mL of potassium bromide solution weigh → 1.03 g.

41.2 mL of potassium bromide solution weigh → ??? g.

∴ The mass of 41.2 mL of potassium bromide solution = (41,2 mL)(1.03 g) / (1.0 mL) = 42.436 g.

∴ The mass of potassium bromide in the solution = the mass of the solution x weight fraction of potassium bromide = (42.436 g) (0.083) = 3.52 g.

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Why is the third equivalence point not observed in the titration curve for phosphoric acid?

BigorU [14]

The third equivalence point is not observed in the titration curve of phosphoric acid because the specific point is concealed due to the rapid ionization of water which in turn forms an hydroxide ion and a molecule called hydronium molecule. The pH value changes more at the first and second points.

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4 years ago

A gas occupies a volume of 180 mL at 35oC and 740 mmHg. What is the volume of the gas at STP?

Tasya [4]

Answer:

there you go

Explanation:

6 0

3 years ago

A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X

barxatty [35]

Answer:

Yes, it is 14. g of compound X in 100 ml of solution.

Explanation:

The relevant fact here is:

the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

84. g solute / 600 ml solution = y / 100 ml solution

⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.

7 0

3 years ago

Add oxidation numbers to the following reaction: 2 H3PO4 (aq) + 2 Cr(s) → 2 CrPO4 (aq) + 3 H2(g). Identify the atom that is oxid

rewona [7]

The atom that is oxidized : Cr

The oxidizing agent : H₃PO₄

<h3>Further explanation</h3>

Reaction

2 H₃PO₄ (aq) + 2Cr(s) → 2 CrPO₄ (aq) + 3H₂(g)

Atoms undergoing a reduction reaction (decrease in oxidation number) and an oxidation reaction (increase in oxidation number)

Reduction (+1 to 0)

H⁺(in H₃PO₄) =+1

H₂=0

Oxidation (0 to +3)

Cr = 0

Cr³⁺(in CrPO₄ )

the oxidizing agent.⇒which undergoes a reduction reaction and oxidizes another compound/element : H₃PO₄

3 0

3 years ago

A food is initially at a moisture content of 90% dry basis. Calculate the moisture content in wet basis

Anvisha [2.4K]

Answer:

Moisture content in wet basis = 47.4 %

Explanation:

Moisture content expresses the amount of water present in a moist sample. Dry basis and wet basis are widely used to express moisture content.

The next equation express the moisture content in wet basis:

$MC_{wb}=\frac{MC_{db}}{1+MC_{db}}$

where, $MC_{wb}$ : moisture content in wet basis and

$MC_{db}$ : moisture content in dry basis

We now calculate the moisture content in wet basis:

$MC_{wb}=\frac{MC_{db}}{1+MC_{db}}$

$MC_{wb}=\frac{0.90}{1+0.90}$

$MC_{wb}$ = 0.474 = 47.4 % wet basis

Have a nice day!

8 0

4 years ago