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2 years ago

PLEASE HELP QUICK

Chemistry

2 answers:

nataly862011 [7]2 years ago

6 0

Answer: Option (a) is the correct answer.

Explanation:

It is known that when an acid is strong then it dissociates completely into the solution of water. Whereas a strong base also dissociates completely into a solution of water.

Also there is little or no reverse reaction of this dissociation.

On the other hand, if an acid or base is not strong enough or weak acid then it will dissociate slightly or partially in a solution of water.

Therefore, when only 42% of a substance dissociates to release hydronium ions then it means it is a weak acid.

wlad13 [49]2 years ago

3 0

A strong acid will dissociate completely in water to release hydronium ions.

This is a weak acid since it is only 42% dissociated.

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The transition metal with the smallest atomic mass

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The transition metal with the smallest atomic mass is Scandium (Sc).
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The noble gases are inert. This means they a. exist as gases at room temperature. b. undergo many chemical reactions. c. lose an

eduard

Answer:

Explanation:

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A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

The final concentration of hydroxide ion = 0.28 M

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