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Studentka2010 [4]
3 years ago
5

A throat spray is 1.40% by mass Phenol, C6H5OH in water. If the solution has a density of 0.9956 g/ml, calculate the molarity of

the solution.
Chemistry
1 answer:
myrzilka [38]3 years ago
8 0

<u>Answer:</u> The molarity of solution is 1.08 M

<u>Explanation:</u>

We are given:

(m/m) of phenol = 1.40 %

This means that 1.40 g of phenol is present in 100 g of solution.

To calculate volume of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 0.9956 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

0.9956g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=100.442mL

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

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (phenol) = 1.40 g

Molar mass of phenol = 94.11 g/mol

Volume of solution = 100.442 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{1.40g\times 1000}{94.11g/mol\times 100.442mL}\\\\\text{Molarity of solution}=0.15M

Hence, the molarity of solution is 0.15 M

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Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at
ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and O_2.

\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

4Al+3O_2\rightarrow 2Al_2O_3

From the balanced reaction we conclude that

As, 3 mole of O_2 react with 4 mole of Al

So, 0.188 moles of O_2 react with \frac{4}{3}\times 0.188=0.251 moles of Al

From this we conclude that, Al is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Al_2O_3

From the reaction, we conclude that

As, 3 mole of O_2 react to give 2 mole of Al_2O_3

So, 0.188 moles of O_2 react to give \frac{2}{3}\times 0.188=0.125 moles of Al_2O_3

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of Al_2O_3 releases amount of heat = 1676 kJ

So, 0.125 mole of Al_2O_3 releases amount of heat = 0.125\times 1676kJ=209.5kJ

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0
3 years ago
We have 548 grams of CaCl2. How many molecules of CaCl2 do we have? (The molar mass of CaCl2 is 110.98 g/mol)
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Answer:

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2.97 x 10^24 molecules

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Explanation:

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gtnhenbr [62]
Hello!

Similarities:

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-They both form Water and Salt when neutralized. 

Differences:

-Acids release Hydronium ions (H₃O⁺) when hydrolyzed while Bases release Hydroxyl ions (OH⁻) when hydrolyzed. 

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4 0
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Answer:

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Hybridization =no. of bond pair +lone pair=3+1=4=sp3 hybridization

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