An experimental procedure says to accurately weigh approximately 1 g of substance an analytical balance. Which of the following

The reaction will yield 2 mol of H₂O. The balanced chemical equation for this process is: c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g). Since 5 moles of O₂ create 4 moles of H₂O, 2.5 mol of O₂ will produce: 2.5 x 4/5 = 2 mol of H₂O.

Answer:

The molality of the solution is 1.08 m.

Explanation:

First, determine the mass of the solvent.

A 13% solution by mass indicates that 13 grams are found in every 100 grams of solution.

Thus, solution mass = solute mass + solvent mass

100 g = 13 g + solvent mass

Therefore, solvent mass = 100 g - 13 g → 87 g

Next, we calculate the moles of solute (mass / molar mass):

13 g / 138.2 g/mol = 0.094 moles

Finally, to find the molality, which is the moles of solute per 1 kg of solvent (mol/kg), we convert the solvent mass to kg:

87 g. 1 kg / 1000 g = 0.087 kg

Then, molality → 0.094 mol / 0.087 kg = 1.08 m

In this scenario, the average cofactor mass for each cell is indicated as 77.91pg, and the total cell quantity is given as 10^5 cells. You need to calculate the total cofactor that will be present in those cells (where 1 microgram equals 10^6 picograms). The calculation follows:
Cofactor mass = cofactor per cell * total cell count = 77.91pg/cell * 10^5 cells = 7.791 x 10^6pg

Next, to convert picograms to micrograms: 7.791 x 10^6pg / (10^6pg/microgram) = 7.791 micrograms

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.

Initially we need to perform the conversion:145 pm = 145 * 10^(-12) m; 36 cm = 360 mm = 360 * 10^(-3) m. Then, we can calculate: 360 * 10^(-3) m / 145 * 10^(-12) m = 360 * 10^(-3) * 10^(12) / 145 = 2.482758621 * 10^(9) or:2,482,758,621 atoms.