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iragen [17]
3 years ago
14

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

ed water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
nadezda [96]3 years ago
5 0

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

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)}}

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL

Putting values in above equation, we get:

2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL

Hence, the volume of NaOH needed is 36.2 mL

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Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel
n200080 [17]

The question is incomplete, here is the complete question:

Consider a mixture of two gases, A and B, confined in a closed vessel. A quantity of a third gas, C, is added to the same vessel at the same temperature. How does the addition of gas C affect the following. The mole fraction of gas B?

A mixture of gases contains 10.25 g of N₂, 2.05 g of H₂, and 7.63 g of NH₃.

<u>Answer:</u> The mole fraction of gas B (hydrogen gas) is 0.557

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For nitrogen gas:</u>

Given mass of nitrogen gas = 10.25 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen gas}=\frac{10.25g}{28g/mol}=0.366mol

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 2.05 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{2.05g}{2g/mol}=1.025mol

  • <u>For ammonia gas:</u>

Given mass of ammonia gas = 7.63 g

Molar mass of ammonia gas = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia gas}=\frac{7.63g}{17g/mol}=0.449mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B+n_C}

Moles of gas B (hydrogen gas) = 1.025 moles

Total moles = [0.366 + 1.025 + 0.449] = 1.84 moles

Putting values in above equation, we get:

\chi_{(H_2)}=\frac{1.025}{1.84}=0.557

Hence, the mole fraction of gas B (hydrogen gas) is 0.557

6 0
3 years ago
Your science teacher gives you three liquids to pour into a jar. After pouring all of them into the jar, the liquids layer as se
sweet-ann [11.9K]

Answer:

density

Explanation:

6 0
3 years ago
What is name of each shape​
dimulka [17.4K]

Answer:

I am so sorry I do not know this

Explanation:

4 0
3 years ago
A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i
Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of KNO_3 in original  254.5 mixture.

moles of BaSO_4 = \frac{mass}{Molecular\ Weight}

moles ofBaSO_4  = \frac{68.3}{233.38}

                               = 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of BaCl_2 = mol\times molecular weight

                         = 0.2926\times 208.23

                         = 60.92 g

the AgCl moles = \frac{mass}{Molecular\ Weight}

                          = \frac{199.1}{143.32}

                          = 1.3891 mol of AgCl

note that, the Cl- derive from both, BACl_2 and NaCl

so

mole of Cl- f NaCl = (1.3891) - (0.2926\times 2) = 0.8039 mol of Cl-

mol of NaCl = 0.8039 moles

mass = mol\times Molecular\ Weight  = 0.8039 \times 58 = 46.626\ g \ of \ NaCl

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

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State the reason that soft drinks is a mixture and water is a compound?
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Carbonated drinks are a mixture of many compounds such as glucose(sugar) and others, while water is simply its own molecule.
7 0
3 years ago
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