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3 years ago

What volume of 24% trichloroacetic acid (tca) is needed to prepare eight 3 ounce bottles of 10% tca solution?

Chemistry

1 answer:

Sever21 [200]3 years ago

6 0

Answer:

295.7 mL of 24% trichloroacetic acid (tca) is needed .

Explanation:

Let the volume of 24% trichloroacetic acid solution be x

Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces

= 24 ounces = 709.68 mL

(1 ounces = 29.57 mL)

Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.

$x\times \frac{24}{100}=709.68 mL\times \frac{10}{100}$

x = 295.7 mL

295.7 mL of 24% trichloroacetic acid (tca) is needed .

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Answer:

The solubility of the gaseous solute decreases

Explanation:

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Hence, in Denver, Colorado where the elevation is about 5,280 feet above sea level, a gaseous solute is less soluble than it is at sea level due to the lower pressure at such high altitude.

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Mayonnaise is made up of eggs, oil, and vinegar. Under which category it should be classified? elements compounds solutions coll

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The answer is 27.92g . working is shown in the picture above.

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How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?

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Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

Water ⇒ C = 1 cal/g°C

m = 65.8 g

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Q = mCΔT

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3 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 28.mg of oxalic acid H2C2

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Answer:

The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

1) Chemical reaction.

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O

b) Balanced chemical equation:

2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

2) Mole ratio

2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

3) Starting amount of oxalic acid

mass = 28 mg = 0.028 g
molar mass = 90.03 g/mol
Convert mass in grams to number of moles, n:

n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol

4) Titration

Volume of base: 28.4 mL = 0.0248 liter
Concentration of base: x (unknwon)

Number of moles of acid: 2.52 mol (calculated above)

Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

$\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622$

That means that there are 0.000622 moles of NaOH (solute)

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M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using three signficant figures, such as the starting data are reported.

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