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Bad White [126]
2 years ago
14

A student is performing a titration to determine the concentration of a 20.0 mL sample of hydrochloric acid. What did the studen

t determine the concentration of the HCl to be if the end point was reached after 50.0 mL of 0.1 M NaOH was added.
Chemistry
1 answer:
irakobra [83]2 years ago
8 0

Answer:

0.25M HCl

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH ⇄ H₂O + NaCl

<em>Where 1 mole of HCl reacts per mole of NaOH</em>

The end point was reached when the student added:

0.0500L × (0.1mol / L) = 0.00500 moles of NaOH

As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:

<em>0.00500 moles HCl</em>

The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:

0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>

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saveliy_v [14]

Answer: 1.98 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}  

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According to stoichiometry :

4 moles of CO_2 will produce =  2 moles of H_2O

Thus 0.22 moles of CO_2 will produce=\frac{2}{4}\times 0.22=0.11moles  of H_2O

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6 0
2 years ago
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mina [271]

Answer:

The diameter of the oil molecule is 4.4674\times 10^{-8} cm .

Explanation:

Mass of the oil drop = m=9.00\times 10^{-7} kg

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Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.

Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m

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Surface of the sphere is given as: a = 4\pi r^2

a=4\times 3.14\times (0.418 m)^2=2.1945 m^2

Volume of the oil drop = v = Area × thickness

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6 0
2 years ago
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<h3>What is Classical bonding?</h3>

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THE ALTERNATIVE IS 4.8g alternative c
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