Lesson Review

Which of the following reagents is/are likely to be used to form the Britton–Robinson buffer solution used in the experiment?

Ierofanga [76]

Answer:

I. A polyprotic, weak acid

II. Na2HPO4

Explanation:

Buffer solutions are those that, upon the addition of an acid or base, are capable of reacting by opposing the part of the basic or acid component to keep the pH fixed.

Buffers consist of hydrolytically active salts that dissolve in water. The ions of these salts are combined with acids and alkalis. These hydrolytically active salts are the products that result from the reaction between weak acids and strong alkalis such as calcium carbonate (from carbonic acid and calcium hydroxide) or between strong acids and weak alkalis such as ammonium chloride (a from hydrochloric acid and ammonium hydroxide).

A buffer acid reacts when a weak acid or weak base is combined with its corresponding hydrolytic salt in a water solution, a buffer system called a buffer is formed. As in this case a weak polyrotic acid with Na2HPO4, which allows the solution to be maintained at a pH of 3.8 against small aggregate amounts of both acid and base, thus favoring the reaction at a pH of 3.8

A buffer system is not always appropriate, because the ions of some hydrolytic salts can, for example, damage organisms that come into contact with it.

4 0

3 years ago

How can density be used to determine the identity of a pure substance?

OLEGan [10]

Answer:

Explanation:

You can identify an unknown substance by measuring its density and comparing your result to a list of known densities. Density = mass/volume. Assume that you have to identify an unknown metal.

5 0

3 years ago

Use the periodic table to correctly describe the arrangement of electrons in an oxygen atom. Check all of the

Anna [14]

Answer:

second option and last option

Explanation:

Know how to do an electron configuration. This is correct because it adds up to 8 and is in the correct order.

All valence electrons are in the last shell. In this case, 2p is the last shell.

6 0

3 years ago

Read 2 more answers

WILL MARK BRAINIEST!!

zloy xaker [14]

Answer:

0 pure reactants

Explanation:

hope this helps!

7 0

3 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago