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MrMuchimi
3 years ago
6

What type of solution forms when lithium chloride dissolved in water?

Chemistry
1 answer:
murzikaleks [220]3 years ago
6 0
Answer: strong electrolyte.

Justification:

1) Lithium chloride is an ionic compound, highly soluble in water.

2)  This is the chemical equation:

LiCl(aq) → Li⁺ (aq) + Cl⁻ (aq)


3) Due to its ionic character and the small size of the ion Li⁺, LiCl dissociates almost completely, meaning the the solution is a strong electrolyte.

4) Electrolytes are solutions that carry electricity (the iones, charged species, are the carrierr of the electricity)

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The correct answer is 12.2% BaO.

The solution is found by dividing the mass of the BaO, which is 25.8 grams, by the total mass of the solution, which is 212 grams, then multiplying it by 100 to get the percentage:
\frac{25.8}{212}*(100) =12.2%
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If the atomic mass of 1 oxygen atom is 15.9994 amu, how much mass does of oxygen have?​
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This question involves two calculations. The answer to the first part will be
ozzi

Answer :

(a) The mass of Al_2O_3 produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of Al = 85.1 g

Molar mass of Al = 27 g/mol

First we have to calculate the moles of Al

\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol

Now we have to calculate the moles of Al_2O_3

The balanced chemical equation is:

4Al+3O_2\rightarrow 2Al_2O_3

From the reaction, we conclude that

As, 4 moles of Al react to give 2 moles of Al_2O_3

So, 3.15 moles of Al react to give \frac{2}{4}\times 3.15=1.58 mole of Al_2O_3

Now we have to calculate the mass of Al_2O_3

\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3

Molar mass of Al_2O_3 = 102 g/mole

\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g

Therefore, the mass of Al_2O_3 produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%

Therefore, the percent yield of the reaction is, 72.5 %

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3 years ago
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