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atroni [7]
3 years ago
5

A 25.0 ml sample of a 0.100 m solution of acetic acid is titrated with a 0.125 m solution of naoh. Calculate the ph of the titra

tion mixture after 10.0
Chemistry
1 answer:
Sedbober [7]3 years ago
5 0

Answer:

1) after 10.0 mL base added  the pH is 4.75

2) after 20.0 mL base added  the pH is  8.74

3)  after 30.0 mL base added  the pH is 12.36

Explanation:

CH₃COOH + NaoH --------------> CH₃COONa + H₂O

1) after 10.0 mL base added

millimoles of NaOH = 10.0 x 0.125 = 1.25

2.50 - 1.25 = 1.25 means 50% titration complete.

at half equivalence point pH = pKa

pH = 4.75

2) millimoles of NaOH added = 20.0 x 0.125 = 2.50

equivalence point

[CH₃COONa] = 2.50 / 20 + 25 = 0.055 M

pH = 1/2 [pKw + pKa + log C]

pH = 1/2 [14 + 4.75 + log 0.055]

pH = 8.74

3) millimoles of NaOH added = 30.0 x 0.125 = 3.75

3.75 - 2.50 = 1.25 millimoles NaOH left

[NaOH] = 1.25 / 30 + 25 = 0.023 M

as NaOH is strong base

[OH-] = [NaOH] = 0.023 M

pOH = - log [OH-]

pOH = - log [0.023]

pOH = 1.64

pH = 14 - 1.64

pH = 12.36

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masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

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3 years ago
Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m
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Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>;  that is

molarity = moles of solute ÷ liters of solution

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

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irina [24]

A.genetic modification of food helps it grow more easily in certain areas

Explanation:

The statement genetic modification of food helps it grow more easily in certain areas is a form of agricultural research that is very accurate.

Genetically modifying food is common practice in order to encourage desirable and sustainable traits in plant species.

  • Drought resistant breeds can tolerate a much lower level of water in some specific areas.
  • Disease resistant breeds are more adapted to outbreaks of plant epidemics.
  • Overall, genetic modification of food helps them to grow and produce better.

learn more:

Genetic recombination brainly.com/question/12685192

#learnwithBrainly

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tiny-mole [99]

Explanation:

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