A solution of HF is titraited with a 0.150M NaOH solution. The pH at the half equivalence point is ? The Ka of HF I 0.00068.

Calculate the theoretical carbonaceous and nitrogenous oxygen demand for:

serg [7]

Answer:

The correct answer is 129 mg and 232 mg.

Explanation:

Theoretical carbonaceous oxygen demand:

The reaction will be,

C₂H₆O₂ + 5/2 O₂ ⇒ 2CO₂ + 3H₂O

Thus, for one mole of C₂H₆O₂ (ethylene glycol), 2.5 moles of O₂ is needed.

The molecular mass of ethylene glycol is 62 grams per mole.

The given mass of ethylene glycol is 100 mg or 0.1 grams

The moles of ethylene glycol will be,

Moles = Weight/Molecular mass

= 0.1/62 = 1.613 × 10⁻³ mol

For 1.613 × 10⁻³ mol, the moles of O₂ will be,

= 2.5×1.613×10⁻³

= 4.0.×10⁻³ × 32mol

= 0.129 grams or 129 mg.

The theoretical nitrogenous oxygen demand is:

The reaction will be,

2NH₃-N + 9/2O₂ ⇒ 4HNO2 + H₂O

Thus, for 2 moles of NH₃-N, 4.5 moles of O₂ is needed,

Therefore, for 1 mol of NH₃-N, the oxygen required will be,

= 4.5/2 = 2.25 mol

The given mass of NH₃-N is 100 mg, the moles of NH₃-N will be,

Moles = 100×10⁻³/31 = 3.225 × 10⁻³ mol (The molecular mass of NH₃-N is 31 gram per mole)

The moles of O₂ is 2.25 × 3.225 × 10⁻³ = 7.258 × 10⁻³ mol.

Now the mass of O2 will be,

= 7.258 × 10⁻³ × 32

= 0.232 grams

= 232 mg

5 0

3 years ago

Naomi has increased the pressure on a solution of liquid and gas in a closed container. What will this do to the gas in her solu

dedylja [7]

Answer:

increase the amount.

Hope this helps =)

Explanation:

5 0

3 years ago

Read 2 more answers

Show the dissolution of KCl, HBr and Methanol via structures also name the type of interaction which helps in its dissolution in

hammer [34]

Answer:

Attached picture

4 0

3 years ago

2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g

uysha [10]

Answer:

The empirical formula is C3H6O

Explanation:

Step 1: Data given

Mass of the sample =2.088 grams

The mass contains carbon, hydrogen, and oxygen

Mass of CO2 produced = 4.746 grams

Mass of H2O produced = 1.943 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 4.746 grams/ 44.01 g/mol

Moles CO2 = 0.1078 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.1078 moles CO2 we'll have 0.1078 moles C

Step 4: Calculate mass C

Mass C: moles C * atomic mass C

Mass C: 0.1078 moles * 12.01 g/mol

Mass C= 1.295 grams

Step 5: Calculate moles H2O

Moles H2O = 1.943 grams / 18.02 g/mol

Moles H2O = 0.1078 moles

Step 6: Calculate moles H

For 1 mol H2O we'll have 2 moles H

For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H

Step 7: Calculate mass H

Mass H = 0.2046 moles * 1.01 g/mol

Mass H = 0.218 grams

Step 8: Calculate mass O

Mass O = 2.088 grams - 1.295 grams - 0.218 grams

Mass O = 0.575 grams

Step 9: Calculate moles O

Moles O = 0.575 grams / 16.0 g/mol

Moles O = 0.0359 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.1078 moles / 0.0359 moles = 3

H: 0.2156 moles / 0.0359 moles = 6

O: 0.0359 moles / 0.0359 moles =1

The empirical formula is C3H6O

8 0

4 years ago

Compute the sugar content in an 8 oz sample of a soft drink. If the sugar content as per label on the product =10g per 100ml.

Nataly_w [17]

Answer:

$m_{sugar}=23.7g\ sugar$

Explanation:

Hello,

In this case, we can first compute the volume of the sample in mL from the ounces:

$8oz*\frac{29.5735mL}{1oz} =236.6mL$

Thus, with the volume of the sample, we can compute the amount of sugar given the 10 g of sugar per 100 mL of soft drink as shown below:

$m_{sugar}=236.6mL*\frac{10g\ sugar}{100mL}\\ \\m_{sugar}=23.7g\ sugar$

Best regards.

6 0

3 years ago