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pentagon [3]
3 years ago
12

Say, for example, that you had prepared a buffer c, in which you mixed 8.203 g of sodium acetate, nac2h3o2, with 100.0 ml of 1.0

m acetic acid. what would be the initial ph of buffer c? if you add 5.0 ml of 0.5 m naoh solution to 20.0 ml each of buffer b and buffer c, which buffer's ph would change less? explain
Chemistry
1 answer:
Alik [6]3 years ago
5 0

Answer:

pH regulatory solution 4.8

pH regulatory solution when adding base 4.86    

Explanation:

Hello!

To calculate the pH of a regulatory solution, the acid constant that relates the acid and the base must be identified in the table. In this case the constant that relates acetic acid and acetate anion has a value Ka = 1.75*10-5.

Assuming that the volume does not vary by aggravating sodium acetate we can calculate its concentration.

Knowing the acid constant and the acid and base concentrations, the pH of the solution is calculated.

By adding a base to the solution, the concentration of the acid species (acetic acid) is decreased by calculating the new concentration, the pH is calculated.

To calculate the concentration after adding the base it is necessary to calculate the dilution of each species, to avoid calculations it is convenient to work with mol or millimol (because they are contained in the same volume can be simplified).

The addition of an acid reduces the formation of the basic species of the solution.

The description of buffer b is missing to compare which pH varies more.

In a larger volume of regulatory solution, lower changes in pH are expected from the addition of acids or bases than in a small volume of solution.

In solutions with equal amounts of acid and base, lower changes in pH are expected than in solutions with different amounts due to the addition of acids or bases.

successes with your homework!  

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love history [14]

1. The molar mass of the unknown gas obtained is 0.096 g/mol

2. The pressure of the oxygen gas in the tank is 1.524 atm

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>1. How to determine the molar mass of the gas </h3>
  • Rate of unknown gas (R₁) = 11.1 mins
  • Rate of H₂ (R₂) = 2.42 mins
  • Molar mass of H₂ (M₂) = 2.02 g/mol
  • Molar mass of unknown gas (M₁) =?

R₁/R₂ = √(M₂/M₁)

11.1 / 2.42 = √(2.02 / M₁)

Square both side

(11.1 / 2.42)² = 2.02 / M₁

Cross multiply

(11.1 / 2.42)² × M₁ = 2.02

Divide both side by (11.1 / 2.42)²

M₁ = 2.02 / (11.1 / 2.42)²

M₁ = 0.096 g/mol

<h3>2. How to determine the pressure of O₂</h3>

From the question given above, the following data were obtained:

  • Volume (V) = 438 L
  • Mass of O₂ = 0.885 kg = 885 g
  • Molar mass of O₂ = 32 g/mol
  • Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
  • Temperature (T) = 21 °C = 21 + 273 = 294 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Pressure (P) =?

The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by V

P = nRT / V

P = (27.65625 × 0.0821 × 294) / 438

P = 1.524 atm

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

Learn more about ideal gas equation:

brainly.com/question/4147359

6 0
3 years ago
How many kPa are in 2,150 mmHg?
LenKa [72]
To determine the pressure in units of kPa, we need to use a conversion factor to convert the units from mmHg to kPa. A conversion factor is a value that would relate two different units and is multiplied or divide to the original measurement depending on what is units is asked. From literature, 1 atm is equal to 760 mmHg and it is also equal to 101.325 kPa. We use these factors to convert the given value. We do as follows:

2150 mmHg ( 1 atm / 760 mmHg ) ( 101.325 kPa / 1 atm ) = 286.643 kPa

Therefore, the closest value from the choices is the second one which has the value of 287, this would be answer.
7 0
3 years ago
2. Calculate the atomic mass of an element that has two isotopes, each with 50.00% abundance. One isotope has a mass of 63.00 am
melamori03 [73]

Answer:

The atomic mass of element is 65.5 amu.

Explanation:

Given data:

Abundance of X-63 = 50.000%

Atomic mass of  X-63 = 63.00 amu

Atomic mass of X-68 = 68.00 amu

Atomic mass of element = ?

Solution:

Abundance of X-68 = 100-50 = 50%

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50×63)+(50×68) /100

Average atomic mass =  3150 + 3400 / 100

Average atomic mass  = 6550 / 100

Average atomic mass = 65.5 amu.

The atomic mass of element is 65.5 amu.

7 0
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Explanation:

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6 0
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Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa
Rashid [163]

Answer:

The temperature is 30,92K

Explanation:

We use the formula PV=nRT. We convert the unit of pressure in kPa into atm.

101,325kPa----1atm

121kPa-------x=(121,3kPax 1 atm)/101,325kPa=1, 2 atm

PV=nRT---->T= (PV)/(RT)

T=(1,2 atm x 3L)/(1,42 mol x 0,082 l atm/K mol )= 30, 91721058 K

6 0
4 years ago
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