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3 years ago

PLEASE ANSWER...........WILL MARK BRAINLIEST!!!!!!!

Chemistry

1 answer:

ANTONII [103]3 years ago

3 0

Answer:

<h2>The Alkali metal halide may precipitate or there may be no change at all</h2>

Explanation:

Alkali metal cations are positively charged. Halogen anions are negatively charged. When a solution of Alkali metal cations is added to a solution of Halogen anions, there are two possibilities :

The alkali metal halide( salt formed from reation of the two ions) may precipitate if the Ionic product is higher than the Solubility product.
However, if it can remain in the solution, it will remain so. No chemical changes happen with respect to these both ions. Nothing willl happen.

There is no reaction happening in either of the cases because both species are already in ionic form before addition, hence they continue to be in this form.

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Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4 What volume of 1.45 M Ca(OH)2 is needed to react with 25.0 moles of H2SO4?

Shkiper50 [21]

Answer:

We need 17.2 L of Ca(OH)2

Explanation:

Step 1: Data given

Concentration of Ca(OH)2 = 1.45 M

Moles of H2SO4 = 25.0 moles

Step 2: The balanced equation

Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4

Step 3: Calculate moles Ca(OH)2

For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4

For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4

Step 4: Calculate volume of Ca(OH)2

Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2

Volume Ca(OH)2 = 25.0 moles / 1.45 M

Volume Ca(OH)2 = 17.2 L

We need 17.2 L of Ca(OH)2

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3 years ago

What is the concentration of OH − and pOH in a 0.00066 M solution of Ba ( OH ) 2 at 25 ∘ C? Assume complete dissociation.

Allushta [10]

<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

<u>Explanation:</u>

We are given:

Concentration of barium hydroxide = 0.00066 M

The chemical equation for the dissociation of barium hydroxide follows:

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

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M$

Putting values in above equation, we get:

$pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88$

Hence, the hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

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Answer:

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