During photosynthesis, sunlight shining on a plant is absorbed. Through several chemical reactions, the plant produces sugar, a

Answer:

(C) The average speed of molecules in ethane is the same as that of propanol.

Explanation:

In gas behavior, temperature is directly linked with speed. At a constant temperature, speed remains consistent. Also, we understand that ideal gases exhibit uniform behavior, irrespective of their type.

1) To express 0.89% m/v, it equals 0.89 grams of NaCl per 100 ml of solution.

This corresponds to 8.9 grams of NaCl in 1000 ml of solution, or 8.9 grams in 1 liter.

2) Molarity is represented as M = moles of solute / liters of solution.

Thus, we need to determine the moles in 8.9 grams of NaCl.

3) The molar mass of NaCl is calculated as 23.0 g/mol + 35.5 g/mol = 58.5 g/mol.

4) Therefore, the number of moles of NaCl calculates as mass / molar mass = 8.9 g / 58.5 g/mol = 0.152 moles.

5) Consequently, M = 0.152 moles of NaCl / 1 liter of solution = 0.152 M.

Answer: 0.152 M

The accurate choice is the final one.

a) The completely balanced chemical reaction is:

 

Zn(s) + H2SO4(aq) --------> ZnSO4(aq) + H2 (g) 

<span>b) Initially, we determine the quantity of zinc that has reacted based on the produced H2.</span>

According to stoichiometry, 1 mole of Zn is required for each mole of H2 created, thus:

moles(Zn) = moles(H2) 

where moles are calculated as the ratio of mass to molar mass (MM)
mass(Zn) / MM(Zn) = mass(H2) / MM(H2) 
mass(Zn) = [mass(H2) / MM(H2)] * MM(Zn) 
mass(Zn) = [(0.0764 g)/(2 g/mol)] * 65.38 g/mol 
mass(Zn) = 2.49 g 

Consequently, we find 2.49 g of pure zinc in the sample, leading to a purity of zinc of: 

purity = (2.49 / 3.86) * 100 % = 64.50 % 
 

<span>c) In part (b), it is assumed that the impurities in the sample do not react with sulfuric acid to emit hydrogen. Thus, the hydrogen solely arises from the reaction of Zn with sulfuric acid.</span>

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.