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Nadusha1986 [10]
3 years ago
10

Given the results of your coarse titration, how does the concentration of the acetic acid compare to the concentration of sodium

hydroxide?
a. The acetic acid solution has the same molarity as the sodium hydroxide solution.
b. The acetic acid solution has a lower molarity than the sodium hydroxide solution.
c. The concentration of the acetic acid solution cannot be determined.
d. The acetic acid solution has a higher molarity than the sodium hydroxide solution.
Chemistry
1 answer:
ratelena [41]3 years ago
5 0

Answer:

d. The acetic acid solution has a higher molarity that the Sodium Hydroxide solution.

Explanation:

Sodium Hydroxide is also known as caustic soda. It is an inorganic compound with molar mass of 39.997 g/mol. It is soluble in water, ethanol and Methanol.  Its formula is NaOH.

Acetic acid is a colorless organic compound liquid called ethanoic acid. It has molar mass of 60.052 g/mol. It is soluble in water. Its formula is CH3COOH.

The molarity of acetic acid is higher than sodium hydroxide.

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1. A gas having the following composition is burnt under a boiler with 50% excess air.
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The composition of the stack gas are :

CH_4= 0.8713

C_3H_8 = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of CH_4 : 70% of 100 g = 70 g

Mass of C_3H_8 : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of CH_4 : \frac{70 g}{16.04 g/mol} = 4.3 mole

Number of moles of C_3H_8: \frac{15 g}{144.1 g/mol} = 0.10 mole

Mass of CO : \frac{15 g}{28.01 g/mol} = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of CH_4: \frac{4.3}{4.935} = 0.8713

Mole fraction of C_3H_8 : \frac{0.10}{4.935} = 0.0202

Mole fraction of CO : \frac{0.53}{4.935} = 0.107

Hence, composition of the stack gas are:

CH_4 = 0.8713

C_3H_8 = 0.0202

CO = 0.107

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