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3 years ago

How many moles of HCI can be formed when 2 mol of hydrogen gas react with chlorine

Chemistry

1 answer:

miss Akunina [59]3 years ago

4 0

So the molar ratio of H2 to HCl is 1:2, so you can make 2mol x 2 = 4 moles of HCl.

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What volume of 0.555M KNO3 solution would contain 12.5 g of solute

Lorico [155]

The volume of 0.555M KNO3 solution would contain 12.5 g of solute iss 223 mL.

<h3>What is the relationship between mass of solute and concentration of solution?</h3>

The mass of solute in a given volume of solution is related by the formula below:

Molarity = mass/(molar mass * volume)

Therefore, volume of solution is given by:

Volume = Mass /molarity * molar mass

Molar mass of KNO₃ = 101 g/mol

Volume = 12.5/(0.555 * 101)

Volume = 0.223 L or 223 mL

In conclusion, the volume of the solution is obtained from the molarity of solution as well as mass and molar mass of solute.

Learn more about molarity and volume at: brainly.com/question/26873446
#SPJ1

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1 year ago

What explains the fact that no machine is 100 percent efficient?

kati45 [8]

Hello, to answer your question. It is because all machines require a fuel or source to power it making it imperfect for using some source that may not be renewable.

Signed by, Virtouso Sargedog

3 0

3 years ago

HELP ME PLEASE I NEED THE ANDEER NOW

AysviL [449]

The answer is 0.430 .

8 0

2 years ago

In a spacecraft tge carbon dioxide exhaled by the astronaut can be removed by the reaction with lithium hydroxide, LiOH accordin

quester [9]

Answer:

30 moles

Explanation:

From the equation it is a one - to - one reaction

3 0

2 years ago

Calcium cyclamate, Ca(C₆H₁₁NHSO₃)₂, is an artificial sweetener used in many countries around the world but is banned in the Unit

xeze [42]

Answer:

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃;

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃;

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

Explanation:

First, let's see the reactants for the first reaction and how they dissociate:

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

BaCO₃ → Ba²⁺ + CO₃²⁻ (Barium is from group 2, so its cation has charge +2)

So, to form the products, the cation of one will join the anion of others. The amount of the cation will be the charge of the anion, and the amount of the anion will be the charge of the cation:

H⁺ + CO₃²⁻ → H₂CO₃

Ba²⁺ + C₆H₁₁NHSO₃⁻ → Ba(C₆H₁₁NHSO₃)₂

The reaction then is:

HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The number of elements must be the same on both sides, so the balanced equation is

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The treatment with H₂SO₄ will produce:

H₂SO₄ → 2H⁺ + SO₄⁻²

Ba(C₆H₁₁NHSO₃)₂ → Ba²⁺ + C₆H₁₁NHSO₃⁻

The balanced reaction will be then:

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃

In the last step, HC₆H₁₁NHSO₃ will react with Ca(OH)₂

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

Ca(OH)₂ → Ca²⁺ + 2OH⁻

The balance reaction will be:

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

3 0

3 years ago