Which of the following rock formations likely existed before the development of a stack?

Answer:

Explanation:

Considering the reaction: 2X + 3Y = 3Z, combining 2.00 moles of X with 2.00 moles of Y results in the production of 1.75 moles of Z.

      2 mol       2 mol   1.75 mol

       2X     +    3Y       =    3Z

2 mol is required with 3 mol to yield 3 mol.

3 mol Z / 3 mol Y =  1 to 1

should yield 2 mol Z

1.75 / 2 = 87.5 % production yield

(a)   Write the balanced half-reactions for the overall process:

Oxidation: Se^2- (aq) → Se (s) + 2e-

Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- → S2O3^2- + 6OH- (aq)

(b)   Assuming E sulfite is 0.57 V, compute E selenium:

E anode = E cathode – E cell

= -0.57 – 0.35

= -.092

Answer:

The configurations are illustrated below.

Explanation:

Hydrogen possesses a single electron in its outer shell, carbon has 4, nitrogen has 5, and oxygen holds 6. To achieve an octet (or duet for hydrogen), they require 1, 4, 3, and 2 electrons respectively.

Therefore, each hydrogen atom will share one electron with carbon, while the remaining electron will be shared with nitrogen, maintaining 4 electrons available for sharing. Carbon can form two bonds with both oxygen atoms, expanding its octet; however, this renders it unstable, leading to the formation of resonance structures (redistribution of electrons), and charge formation. One of the oxygen atoms will share only one electron with nitrogen.

The two structures are depicted below.

Answer:

The percent yield of Br₂ in this reaction amounts to 96.15%

Explanation:

The reaction's balanced stoichiometric equation is:

2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

To calculate the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

To determine the theoretical yield:

5.29 g of NaBr reacts with an excess of chlorine; therefore, NaBr is the limiting reagent, controlling the possible yield of products.

We convert 5.29 g of NaBr to moles.

Number of moles = (Mass)/(Molar mass)

Molar Mass of NaBr = 102.894 g/mol

Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole

According to the stoichiometry of the reaction:

2 moles of NaBr yield 1 mole of Br₂

Thus, 0.05141 mole of NaBr will produce (0.05141×1/2) mole of Br₂, which is 0.0257 mole of Br₂

Theoretical yield = Expected mass of Br₂ from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Calculating the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

Actual yield = 3.95 g

Theoretical yield = 4.108 g

Percent yield = 100% × (3.95/4.108) = 96.15%

Hope this is helpful!!!