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2 years ago

Calculate the pH of a buffer solution prepared by mixing 75.0 mL of 1.00 M lactic acid and 25.0 mL of 0.50 M sodium lactate.

Chemistry

1 answer:

max2010maxim [7]2 years ago

6 0

From the calculations, the pH of the buffer is 3.1.

<h3>What is the pH of the buffer solution?</h3>

The Henderson-Hasselbach equation comes in handy when we deal with the pH of a buffer solution. From that equation;

pH = pKa + log[(salt/acid]

Amount of the salt = 25/1000 * 0.50 M = 0.0125 moles

Amount of the acid = 75/1000 * 1.00 M = 0.075 moles

Total volume = ( 25 + 75)/1000 = 0.1 L

Molarity of salt = 0.0125 moles/0.1 L = 0.125 M

Molarity of the acid = 0.075 moles/0.1 L = 0.75 M

Given that the pKa of lactic acid is 3.86

pH = 3.86 + log( 0.125/0.75)

pH = 3.1

Learn more about pH:brainly.com/question/5102027

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The rate constant for the decomposition of nitrogen dioxide NO2(g) LaTeX: \longrightarrow⟶ NO (g) + 1/2 O2(g) with a laser beam

aleksley [76]

Answer :

The time taken by the reaction is 19.2 seconds.

The order of reaction is, second order reaction.

Explanation :

The general formula to determine the unit of rate constant is:

$(Concentration)^{(1-n)}(Time)^{-1}$

Unit of rate constant Order of reaction

$Concentration/Time$ 0

$Time^{-1}$ 1

$(Concentration)^{-1}Time^{-1}$ 2

As the unit of rate constant is $M^{-1}min^{-1}$ . So, the order of reaction is second order.

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $1.95M^{-1}s^{-1}$

t = time = ?

$[A_t]$ = final concentration = 0.97 M

$[A_o]$ = initial concentration = 2.48 M

Now put all the given values in the above expression, we get:

$1.95\times t=\frac{1}{0.97}-\frac{1}{2.48}$

$t=0.32min=0.32\times 60s=19.2s$

Therefore, the time taken by the reaction is 19.2 seconds.

8 0

3 years ago

How many particles are in a 151 g sample of Li2O?

neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0

3 years ago

Use a periodic table to help you answer these questions if needed.

yuradex [85]

Answer:

2 Al and 3 S

Explanation:

Al2S3 so 2 Al and 3 S

3 0

1 year ago

As you were climbing a a mountain, you noticed that rocks were crumbling below your feet and moving down the mountain. What two

Flauer [41]

Answer:

Weathering and erosion

Explanation:

Weathering can be explained as the breaking down of rocks/minerals on the surface of the Earth as a result of contact with biological organism, water, air and other factors

. There are 3 common types of weathering which are;

1) physical weathering

2) biological weathering

3) chemical weathering

Erosion can be regarded as a geological process, whereby earthen material are been transported away by natural forces, these forces could be wind as well as water.

Therefore, as you were climbing a a mountain, you noticed that rocks were crumbling below your feet and moving down the mountain. What is observed are weathering and erosion processes.

After the weakening and broken up of the rock by weathering then erosion transport the bit of the rock down the mountain as you are climbing, which means the "weathering process" breakdown and the "erosion process" involves the transport or movement of the bit of the rocks

6 0

3 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago