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AfilCa [17]
2 years ago
9

Suppose you carry out a titration involving 1.90 molar CsOH and an unknown concentration of HI. To bring the reaction to its end

point, you add 9.9 milliliters of CsOH to 25.0 milliliters of HI.
Now you know that you have a _ solution of HI.
Chemistry
1 answer:
Mariulka [41]2 years ago
3 0

The concentration of the HI solution is 0.75M.

<h3>How do we calculate the required concentration?</h3>

Required concentration of any solution used in titration will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ are the molarity and volume of CsOH.

M₂ & V₂ are the molarity and volume of HI.

On putting all values from the question, we get

M₂ = (1.9)(9.9) / (25) = 0.75M

Hence required concentration of HI solution is 0.75M.

To know more about concentration & volume, visit the below link:
brainly.com/question/24697661

#SPJ1

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When substances react chemically to create a new substance, will that new substance have the same physical and chemical properti
skad [1K]

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No it will not have the same chemical properties as the initial substance.

Explanation:

In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants

4 0
3 years ago
Two solutions are created by mixing one solution containing lithium nitrate with one containing sodium phosphate.
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Answer:

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

Explanation:

                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

Where S = Solubility(mole/lit) and Ksp = Solubility product

⇒ Ksp = (3S)³ x (S)

⇒ 27S⁴ = 2.3x10−4

⇒ S = 0.05 mol/lit

Concentration of Li₃PO₄ precipitate = 0.05

<u>Solution A </u>

0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole

0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)    \frac{0.15}{3}                \frac{0.24}{1}

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Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

                       

6 0
3 years ago
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Hi there! Let's solve this problem shall we!

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In this specific problem, they are asking us to find the <u><em>density </em></u>of the object. So,<u><em> using the information given to us</em></u> (volume and mass), let's solve the problem!

Now, if you remember, D = M ÷ V

So, let's fill in the blanks!

D = Our unknown value

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V = 10g

Here is the filled out formula:

D = M ÷ V

D = 2mL ÷ 10g

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*Make sure you put the units for your final solution!*

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