When the volume of gas is changed from 3.75 to 6.52 L , the temperature will change from 100.0 K to how many k?

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol / 2 mol CH3COOH)

= 0.005 mol

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm

Step 1: Convert density from g/mL to g/L; 0.807 g/mL is equivalent to 807 g/L. Step 2: Calculate Moles of N₂; Density = Mass / Volume, or Mass = Density × Volume. Plugging in values, Mass = 807 g/L × 1 L gives us Mass = 807 g. Similarly, Moles = Mass / M.mass, which leads to Moles = 807 g / 28 g.mol⁻¹, giving us Moles = 28.82 moles. Step 3: Apply the Ideal Gas Law to determine Volume of gas occupied; P V = n R T, thus V = n R T / P. Remember to convert temperature to Kelvin (25 °C + 273 = 298 K). Hence, V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm, resulting in V = 704.76 L.

Please ask if you have any inquiries.

Answer:

Endothermic: water formation from ice and a cold instant ice pack  Explanation:

Response:

a. 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

b. Br₂

c. Br₂

Clarification:

Balancing a redox reactionis performed using the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: Br₂(l) → Br⁻(aq)

Oxidation: Br₂(l) → BrO₃⁻(aq)

Step 2: Perform mass balance. This reaction occurs in basic conditions, thus we must add OH⁻ and H₂O as needed.

0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O

Step 3: Ensure electrical balance by incorporating electrons when necessary.

1 e⁻ + 0.5 Br₂(l) → Br⁻(aq)

6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻

Step 4: Scale both half-reactions to ensure the electron counts balance out.

5 × (1 e⁻ + 0.5 Br₂(l) → Br⁻(aq))

1 × (6 OH⁻(aq) + 0.5 Br₂(l) → BrO₃⁻(aq) + 3 H₂O + 5 e⁻)

Step 5: Combine both half-reactions and simplify as appropriate.

5 e⁻ + 3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O + 5 e⁻

3 Br₂(l) + 6 OH⁻(aq) → 5 Br⁻(aq) + BrO₃⁻(aq) + 3 H₂O

The species that undergoes reduction is identified as the oxidizer. The species that undergoes oxidation is termed the reducer. In this situation, Br₂ qualifies as both.