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A buffer solution contains 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. What is the pH of this buffer?What is the pH of the buffer after the addition of 2?

Answers

Answer 1

Answer:

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76

(a). pH = 4.76 + log [0.13]/[0.10]

= 4.76 + 0.11

= 4.87

(b) KOH + CH3COOH =>H2O + CH3COOK so (acid)goes down and (salt)goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt

pH = 4.76 + log [0.15]/[0.08]

= 4.76 + 0.27

= 5.03

Answer 2

Answer:

Final answer:

The pH of the buffer with 0.11 mol acetic acid and 0.13 mol sodium acetate in 1.00 L is 4.91, calculated using the Henderson-Hasselbalch equation. The second part of the question regarding the pH change after the addition of '2' is unanswerable without further information on what is being added.

Explanation:

To answer the question of what the pH of the buffer solution containing 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L is, we can apply the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. For acetic acid, the pKa is approximately 4.76. Since we have 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L solution, the concentrations are 0.11 M and 0.13 M respectively.

Substituting these values into the Henderson-Hasselbalch equation gives:

pH = 4.76 + log(0.13/0.11)

Calculating the log(0.13/0.11) yields approximately 0.15. Therefore:

pH = 4.76 + 0.15 = 4.91

The question "What is the pH of the buffer after the addition of 2?" seems to be incomplete, as it does not specify what '2' refers to. If '2' refers to adding 2 moles of a strong acid or base, for instance, the pH would change significantly and the buffer capacity might be exceeded. The exact effect on pH would depend on the nature of the substance added (acid or base) and its quantity. Without specifics, this part of the question cannot be accurately answered.

The concept of buffer capacity is relevant to discuss here. Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant change in pH. Buffer solutions with higher molar concentrations of both the acid and the corresponding salt will have greater buffer capacity.

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A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 protons and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay

Answers

Answer:

No additional particle was produced during the decay.

Explanation:

The equation of decay is given as;

¹⁰₆C + ⁰₋₁ e → ¹⁰₅B + x

To identify x, we have to calculate its atomic and mass number.

In the reactants side;

Atomic Number = 6 + (-1) = 5

Mass number = 10 + 0 = 10

In the products side;

Atomic Number = 5 + x

Mass Number = 10 + x

Generally, reactant = product

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Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressures in atmospheres to three significant digits separated by commas.

Answers

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of $O_2$ and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = $n_1=(51.8)/(32 g/mol)=1.619 mol$

Moles of helium gas = $n_2=(33.1 g)/(4 g/mol)=8.275 mol$

Total moles of gas = $n_1+n_2=(1.619 +8.275 ) mole=9.894 mol$

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

$P=(nRT)/(V)$

$=(9.894 mol* 0.0821 atm L/mol K* 294 K)/(10.0 L)$

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

$p_(1)=P* \chi_(1)=P* (n_1)/(n_1+n_2)$

$p_1=23.88 atm* (1.619 mol)/(9.894 mol)=3.91 atm$

Partial pressure of the helium gas

$p_(2)=P* \chi_(2)=P* (n_2)/(n_1+n_2)$

$p_2=23.88 atm* (8.275 mol)/(9.894 mol)=19.97 atm\approx 20.0 atm$

What is anology?
What is microscope?

Answers

Microphone thing is a little bit better but it was ok I gotta it looked good but it didn’t get me wrong it I made a comment on my phone number so ttyttttytt was the day I wanna was a good day and then I’ll send ya my stuff and I get to see ya my mom stuff is the best thing ever I just wanna was a

How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? a. 37.50 mL
b. 50.00 mL
c. 75.00 mL
d. 100.00 mL
e. 25.00 mL

Answers

Answer:

We need 75 mL of 0.1 M NaOH ( Option C)

Explanation:

Step 1: Data given

Molarity of NaOH solution = 0.100 M

volume of 0.150 M CH3COOH = 50.00 mL = 0.05 L

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles of CH3COOH

Moles CH3COOH = Molarity * volume

Moles CH3COOH = 0.150 M * 0.05 L

Moles CH3COOH = 0.0075 moles

Step 4: Calculate moles of NaOH

For 1 mol of CH3COOH we need 1 mol of NaOH

For 0.0075 mol CH3COOH we need 0.0075 mole of NaOH

Step 5: Calculate volume of NaOH

volume = moles / molarity

volume = 0.0075 moles / 0.100 M

Volume = 0.075 L = 75 mL

We need 75 mL of 0.1 M NaOH

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Answers

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

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Now,

Increase in volume will be:

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Which of the following is an example of physical change?

Answers

Answer:

The glass cup falling from the counter

Explanation:

the glass isn't changing in any chemical way. it's still made of the same material, just broken apart.

Final answer:

Physical changes involve the alteration of the state or appearance of matter, without changing the composition. An example is solid wax turning into liquid wax when heated, or steam condensing inside a cooking pot.

Explanation:

The question asks for an example of a physical change. Physical changes involve alterations in the state or appearance of matter, without changing its composition. For example, solid wax turning into liquid wax when heated is a physical change. The wax is still the same substance, it's just in a different state. Similarly, steam condensing inside a cooking pot is also a physical change. The water vapor turns back into liquid water, but it's still water. These are distinguished from chemical changes, which transform one substance into a different substance.

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