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asambeis [7]
3 years ago
11

A student performs three trials of a titration of an acid with an unknown concentration. She compares her measured concentration

to the actual value, and calculates a 25% error. Assuming she has performed all of the calculations correctly, name one aspect of her methods that may have led to such a high error.
Chemistry
1 answer:
Viktor [21]3 years ago
4 0

Answer:

There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.

Main error that leads to the error in results is misreading of the end point volume .

End point is when the reaction between the analyte and solution of known concentration has stopped .

Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .

From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.

So, error may have occurred in wrongly judging of the end point by color change of the indicator .

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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?
natita [175]

This question provides us –

  • Weight of \bf  KCl is = 47 g
  • Volume, V = 375 mL

__________________________________________

  • Molar Mass of \bf   KCl –

\qquad \twoheadrightarrow\bf  39.0983 \times 35.453

\qquad \twoheadrightarrow\bf 74.5513

<u>Using formula</u> –

\qquad \purple{\twoheadrightarrow\bf Molarity _{(Solution)} =  \dfrac{ W\times 1000}{MV}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{ 47 \times 1000}{74.5513\times 375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{47000}{27956.7375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \cancel{\dfrac{47000}{27956.7375}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = 1.68117M

\qquad \pink{\twoheadrightarrow\bf Molarity _{(Solution)}  = 1.7M}

  • Henceforth, Molarity of the solution is = 1.7M

___________________________________________

6 0
2 years ago
Which is NOT an example of a solution?
USPshnik [31]
C table snap Lt it is a compound
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The Ka of carbonic acid is 4.3 x 10-7.
ratelena [41]

Answer:

poor hydrogen-ion donor

Explanation:

Acid dissociation constant constant chemistry is the equilibrium constant of the dissociation reaction of an acid, it is denoted by Ka. This equilibrium constant is a measure of the strength of an acid in a solution.

Note these as a rule of thumb:

When Ka is large, the dissociation of the acid is favored.

When Ka is small, the acid does not dissociate to a large extent.

Hence, a Ka of 4.3 x 10-7 shows a weak acid. A weak acid is a poor hydrogen ion donor because it does not dissociate to a large extent in solution.

3 0
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Talja [164]

Answer:

I dont know

Explanation:

coz I don't know the answer

7 0
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When heated, solid copper(II) carbonate decomposes to solid copper(II) oxide and carbon dioxide gas. Give the balanced chemical
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Answer:

Explanation:

CuCO3(s) ====> Cu(ii)O(s) + CO2(g)

Nothing else needs to be done. The equation is balanced.

8 0
2 years ago
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