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ivolga24 [154]
3 years ago
14

If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con

centration of hydroxide (OH-) ions in the 25.00 mL titrated sample of Ca(OH)2 solution.
Chemistry
1 answer:
drek231 [11]3 years ago
6 0

Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

The balanced equation of reaction is:

2HCl + Ca(OH)2 ===> CaCl2 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = 1M

CB is the concentration of base = to be calculated

VA is the volume of acid = 23.65 ml

VB is the volume of base = 25mL

Substituting

1×23.65/CB×25 = 2/1

Therefore CB =1×23.65×1/25×2

CB = 0.473 M.

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Answer:

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Explanation:

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How should you measure the mass of a given powdered substance??
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How many atoms are in 2.12 mole of propane
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A single hydrogen atom has a mass of 1.67 × 10−24 g. A sodium atom has an atomic mass of 23. How many sodium atoms are required
statuscvo [17]
<h3>Answer:</h3>

                      2.55 × 10²² Na Atoms

<h3>Solution:</h3>

Data Given:

                 M.Mass of Na  =   23 g.mol⁻¹

                 Mass of Na  =  973 mg  =  0.973 g

                 # of Na Atoms  =  ??

Step 1: Calculate Moles of Na as:

               Moles  =  Mass ÷ M.Mass

               Moles  =  0.973 g ÷ 23 g.mol⁻¹

               Moles  =  0.0423 mol

Step 2: Calculate No, of Na Atoms as:

As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,

               Moles  =  No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹

Solving for No. of Na Atoms,

               No. of Na Atoms  =   Moles × 6.022 × 10²³ Na Atoms.mol⁻¹

               No. of Na Atoms  =   0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹

               No. of Na Atoms  =  2.55 × 10²² Na Atoms

<h3>Conclusion: </h3>

                          2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.


5 0
3 years ago
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