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UkoKoshka [18]
3 years ago
11

How many moles of HBr are present in 250 ml of a 0.15 M HBr solution?

Chemistry
1 answer:
Fofino [41]3 years ago
6 0

Answer:

0.0375 moles HBr

Explanation:

we are given;

  • Molarity of HBr solution as 0.15 M
  • Volume of the solution as 250 mL

We are required to determine the number of moles;

We need to know that;

Molarity = Moles ÷ Volume

Therefore;

Moles of HBr = Molarity of HBr solution × Volume of solution

Thus;

Moles of HBr = 0.15 M × 0.25 L

                      = 0.0375 Moles

Thus, the number of moles of HBr solute is 0.0375 moles

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Why can't chemical changes be undone? Because of the chemical properties. Because Mark is an idiot. Because they need energy. Be
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Answer: Because new substances are formed.

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What are the empirical formula and empirical formula mass for C10H30O10?
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Answer:

Empirical formula: CH₃O

Empirical formula mass = 31 g/mol

Explanation:

Data Given:

Molecular Formula = C₁₀H₃₀O₁₀

Empirical Formula = ?

Empirical Formula mass =

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule

As

C₁₀H₃₀O₁₀ Consist of  10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.

Now

Look at the ratio of these three atoms in the compound

                         C : H : O

                        10 : 30 : 10

Divide the ratio by two to get simplest ratio

                          C      :   H      :    O

                         10/10 : 30/10 : 10/10

                             1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

                     Empirical formula of C₁₀H₃₀O₁₀ =  CH₃O

Now

To find the empirical formula mass in g/mol

Formula mass:

Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.

**Note:

if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol

So,

As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O

Then Its empirical formula mass will be

CH₃O

Atomic Mass of C = 12

Atomic Mass of H = 3

Atomic Mass of O = 16

Total Molar mass of CH₃O

CH₃O = 12 + 3(1) + 16

CH₃O = 12 + 3 + 16

CH₃O = 31 g/mol

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