Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.700 atm

Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.

Answer:

a. H₂O (conjugate acid); b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid); c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base); d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base); f. No conjugate acid or base exists; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide functions as a Bronsted-Lowry base, allowing it to capture a proton, thus water serves as the conjugate acid.

b. H₂O is amphoteric, capable of acting as either an acid or a base. As a base, its conjugate acid is H₃O⁺, whereas as an acid, its conjugate base is OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

Bicarbonate is also amphoteric. When it captures a proton, it forms carbonic acid as the conjugate acid when acting as a base. When HCO₃⁻ acts as an acid and releases a proton, carbonate becomes the conjugate base.

d. Ammonia functions as a weak base, with ammonium being the conjugate strong acid.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. Acid sulfate can function as both an acid and a base.

(similar to bicarbonate). Acting as a base yields sulfuric acid as the conjugate acid, while acting as an acid leads to sulfate as the conjugate base.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not accept H⁺ or OH⁻ nor does it expel H⁺. It’s neutral and does not function as an acid or base.

g. HS⁻ is amphoteric.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

This is similar to the case of bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hydrazinium acts as an acid, making hydrazine its conjugate base.

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

The response is:

No, the equation is not balanced. Neither the Nitrogen (N) nor the Hydrogen (H) are in balance!

Here's the reasoning:

⓵ A properly balanced chemical equation means that the quantity of atoms on the reactants side matches that on the products side.

→ The equation lacks balance because there are 2 Nitrogen atoms and 2 Hydrogen atoms on the reactants side. In contrast, on the products side, there is only 1 Nitrogen atom and 4 Hydrogen atoms. Thus, the number of atoms on each side is not consistent!

Hopefully, this clarification is helpful; feel free to reach out if you have any further questions! ☻