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Galina-37 [17]
3 years ago
10

A cough syrup contains 5.0 % ethyl alcohol C2H5Oh by mass. If the density of the solution is 0.9928 g/ml. Determine the molarity

of the alcohol in the cough syrup.
Chemistry
1 answer:
Nostrana [21]3 years ago
6 0

<u>Answer:</u> The molarity of ethyl alcohol in cough syrup is 1.08 M

<u>Explanation:</u>

We are given:

(m/m) of ethyl alcohol = 5.0 %

This means that 5 g of ethyl alcohol 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 = 0.9928 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

0.9928g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=100.725mL

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 (ethyl alcohol) = 5 g

Molar mass of ethyl alcohol = 46.07 g/mol

Volume of solution = 100.725 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{5g\times 1000}{46.07g/mol\times 100.725mL}\\\\\text{Molarity of solution}=1.08M

Hence, the molarity of ethyl alcohol in cough syrup is 1.08 M

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The following list contains some common radicals. Using the charges on these ions and the idea of valence, predict the formulas
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Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

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Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to
Natalija [7]

Answer:

1- 3 Moles of CO2  

2- 132 g of CO2  

3- 105,6 g of CO2

4- Limiting Reagent O2

<u>Products form based on limiting reagent (384g O2) :</u>

CO2: 316,8 g

H2O: 172,8 g

<u>Products form based on C3H8 (132,33 g):</u>

CO2: 396,99 g

H2O: 216,54 g  

Explanation:

<u>Atomic Masses:</u>

C: 12

H: 1

O: 16

<u>Molecular weights:</u>

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g  O2 ⇒ 132 g CO2

128 g  O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For          44 g of CH3H8  ⇒160 g of O2

With   132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

<u>Calculations of the products formed in base of quantity of O2 (limiting reagent):</u>

160 g  O2 ⇒ 132 g CO2

384 g  O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g  O2 ⇒ 72 g H2O

384 g  O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

<u>Calculations of the products formed in base of quantity of C3H8 (excess reagent):</u>

     44 g  C3H8 ⇒ 132 g CO2

132,33 g  C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

     44 g C3H8 ⇒ 72 g H2O

132,33 g  C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

<u />

3 0
3 years ago
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