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olasank [31]
2 years ago
10

A 19.5-mL sample of Ba(OH)2 is titrated with HCl. If 28.4 mL of 0.394 M HCl is needed to reach the endpoint, what is the concent

ration (M) of the Ba(OH)2 solution
Chemistry
1 answer:
Romashka-Z-Leto [24]2 years ago
5 0

The concentration of Ba(OH)₂ IS 0.287 M.

<h3>What is concentration?</h3>

This is the amount of solute in mol or in gram that will dissolve in 1 dm³ solution.

Equation of the reaction

  • Ba(OH)₂+2HCl→ BaCl₂+2H₂O

Formula:

  • CaVa/CbVb = Na/Nb................ Equation 1

Where:

  • Ca = Concentration of HCl
  • Va = Volume of HCl
  • Cb = Concentration of Ba(OH)₂
  • Vb = Volume of Ba(OH)₂
  • Na = number of moles of HCl
  • Nb = number of moles of Ba(OH)₂

Make Cb the subject of the equation

  • Cb = (CaVaNb)/(NaVb)................... Equation 2

From the equation and the question,

Given:

  • Ca = 0.394 M
  • Va = 28.4 mL
  • Vb = 19.5 mL
  • Na = 2
  • Nb = 1

Substitute these values into equation 2

  • Cb = (0.394×28.4×1)/(2×19.5)
  • Cb = 0.287 M.

Hence, The concentration of Ba(OH)₂ IS 0.287 M.

Learn more about concentration here: brainly.com/question/17206790


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An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as \pi* orbitals.

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Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO
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NO would form 65.7 g.

H₂O would form 59.13 g.

Explanation:

Given data:

Moles of NH₃ = 2.19

Moles of O₂ = 4.93

Mass of NO produced = ?

Mass of  produced H₂O = ?

Solution:

First of all we will write the balance chemical equation,

4NH₃ + 5O₂   →   4NO + 6H₂O

Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:

NH₃  :   NO                                   NH₃  :   H₂O

4     :    4                                          4    :      6

2.19   :    2.19                                 2.19  : 6/4 × 2.19 = 3.285 mol

Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂  :   NO                                               O₂ :   H₂O

5     :    4                                                  5     :    6

4.93   :   4/5×4.93 = 3.944 mol               4.93  : 6/5 × 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of  moles × molar mass

Mass of water = 3.285 mol × 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide  = number of  moles × molar mass

Mass of nitrogen monoxide = 2.19 mol × 30 g/mol

Mass of nitrogen monoxide = 65.7 g

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