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777dan777 [17]
3 years ago
7

Calculate the mass of 5.3 moles of chloric acid (HClO3). Please Answer ASAP!!!! :((((((\

Chemistry
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

292.215

Explanation:

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What is the name of B2(SeO4)3
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This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.

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C₇H₆O₂ + O₂ -->CO₂ +H₂O Find the chemical reaction
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Assuming 100% dissociation, calculate the freezing point and boiling point (in °C) of a solution with 83.6 g of AgNO 3 in 1.00 k
Thepotemich [5.8K]

Answer:

  • <u>Freezing point: - 1.83ºC</u>
  • <u>Boiling point: 100.50ºC</u>

Explanation:

The <em>freezing point</em> and<em> boiling point</em> of solvents, when a solute is added, will change accordingly to the concentration of the solute particles.

The freezing point will decrease and the boiling point will increase. These are two colligative properties.

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3 0
3 years ago
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
3 years ago
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