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andriy [413]
3 years ago
14

Determine the volume in mL of 0.37 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 2

0.8 mL of 0.51 M CH3CH2NH2(aq). Enter your answer with one decimal place. The Kb of ethylamine is 6.5 x 10-4.
Chemistry
1 answer:
Dimas [21]3 years ago
7 0

Answer:

14.3mL you require to reach the half-equivalence point

Explanation:

A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:

CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.

Initial moles of CH₃CH₂NH₂ are:

20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =

0.0106moles CH₃CH₂NH₂

To reach the half-equivalence point you require:

0.0106moles ÷ 2 = 0.005304 moles HClO₄

As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:

0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =

<h3> 14.3mL you require to reach the half-equivalence point</h3>

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Harlamova29_29 [7]

Answer: 2.62 x 10^22 atoms

Explanation:Please see attachment for explanation

5 0
2 years ago
Write the chemical formulae of<br>a)dinitrogen oxide<br>b) calcium dihydrogen phosphate​
il63 [147K]

Answer:

A) <u><em>Dinitrogen Oxide:</em></u>

=> N_{2}O (Also called Nitrous Oxide)

B) <u><em>Calcium Dihydrogen Phosphate​:</em></u>

=> Ca(H_{2}PO_{4})_{2} (Also called Monocalcium phosphate)

4 0
3 years ago
A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.
lord [1]

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g

The produced energy will be:

=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K

=450.35\times 3.2

=1441.12 \ J

The reaction will be:

⇒  HCl+NaOH \rightarrow NaCl+H_{2}O

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.  

Therefore, the moles generated by NaCl will indeed be:

=  (\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}

=  0.0318\times 0.500

=  0.0159 \ mole  \ of \ NaCl

Now,

=  \frac{1441.12 \ J}{0.0159 \ moles \ NaCl}

=  906364.7

=  90.6 \ KJ/mol \ NaCl

7 0
2 years ago
What is the molarity of a HCl solution, if 28.6 mL of a 0.175 m NaOH solution is needed to neutralize a 25.0 mL sample of the HC
Katen [24]

The molarity of the HCl solution needed to neutralize 28.6 mL of a 0.175 M NaOH solution is 0.2002 M

We'll begin by writing the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

  • From the question given above, the following data were obtained:

Volume of base, NaOH (Vb) = 28.6 mL

Molarity of base, NaOH (Mb) = 0.175 M

Volume of acid, HCl (Va) = 25 mL

<h3>Molarity of acid, HCl (Ma) = ?</h3>

The molarity of the acid, HCl can be obtained as follow:

MaVa / MbVb = nA / nB

(Ma × 25) / (0.175 × 28.6) = 1

(Ma × 25) / 5.005 = 1

Cross multiply

Ma × 25 = 5.005

Divide both side by 25

Ma = 5.005 / 25

<h3>Ma = 0.2002 M</h3>

Therefore, the molarity of the acid, HCl needed for the reaction is 0.2002 M

Learn more: brainly.com/question/25573711

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