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3 years ago

How much of a 0.250 M lithium hydroxide is required to neutralize 20.0 mL of 0.345M chlorous acid?

Chemistry

1 answer:

Bumek [7]3 years ago

6 0

Answer:

27.6mL of LiOH 0.250M

Explanation:

The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:

LiOH + HClO₂ → LiClO₂ + H₂O

That means, 1 mole of hydroxide reacts per mole of acid

Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:

0.0200L ₓ (0.345mol / L) = 6.90x10⁻³ moles of HClO₂

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

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Answer:34.55
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3 years ago

The mass of an object is 240g and its volume is 60 cm3 A. Sink B. Float

Vinil7 [7]

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Explanation:

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4 years ago

S SIO₂ +3C Sic +20o Sic ? mass in grams С c-

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Answer:

26.74g

Explanation:

The equation of the reaction is;

SIO₂ + 3C --> SiC +2CO

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3 mol of C produces 1 mol of SiC

Converting mol to mass using; Mass = moles * Molar mass

Mass of SiC = 1 mol * 40.11 g/mol = 40.11g

This means;

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6 0

3 years ago

Pls give me the right answer

Alchen [17]

The answer to the question is D

5 0

2 years ago

A student creates a solution by dissolving some Na2SO4 (molar mass = 142.05 g/mol) in enough water to create 0.500 L of solution

NeX [460]

Answer : The mass of $Na_2SO_4$ is, 4.43 grams.

Explanation :

As we know that, when $Na_2SO_4$ dissolve in water then it dissociates to give 2 mole of sodium ion $Na^+$ and 1 mole of sulfate ion $SO_4^{2-}$

The chemical reaction will be:

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

Thus, the concentration of $Na_2SO_4$ = $\frac{\text{Concentration of }Na^+}{2}=\frac{0.125M}{2}=0.0625M$

First we have to calculate the moles of $Na_2SO_4$

$\text{Concentration of }Na_2SO_4=\frac{\text{Moles of }Na_2SO_4}{\text{Volume of solution}}$

$0.0625M=\frac{\text{Moles of }Na_2SO_4}{0.500L}$

$\text{Moles of }Na_2SO_4=0.0312mol$

Now we have to calculate the mass of $Na_2SO_4$

$\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4$

Molar mass of $Na_2SO_4$ = 142 g/mol

$\text{Mass of }Na_2SO_4=0.0312mol\times 142g/mol=4.43g$

Thus, the mass of $Na_2SO_4$ is, 4.43 grams.

6 0

3 years ago