Which term describes the difference in electrical charge across a membrane?

The oxidation state numbercan aid in identifying the unknown element present in both compounds. They denote the number of electrons that are either donated, received, or shared to yield compounds.

Remember the fundamental principles governing oxidation numbers.

1. In a neutral compound, the total of all oxidation numbers is zero.
2. Chlorine, bromine, and iodine typically exhibit an oxidation number of -1(unless paired with fluorine and oxygen)

Assume the oxidation state for element Mis designated as x.
Referring to rule 2, chlorine possesses an oxidation state of -1.

Now, for the compound MCl₂ (which is neutral), the equation can be formulated as
x + (2 * -1)= 0 ⇒ x₁= +2

For MCl₃, the corresponding equation is
x + (3 * -1)=  0 ⇒ x₂= +3

This indicates that the elementhas two distinct oxidation statesin its compounds, which are +2and +3.

The identified element is iron (Fe), as it shows +2 and +3 oxidation states across these compounds.

Memorizing this is essential. Regrettably, there isn't a simpler method to tackle these oxidation states.

The final answer is iron (Fe).

Answer:

Ir(NO2)3

Explanation:

The molar mass is 330.2335, in case that's also required.

The L- isomer serves as the enantiomer of the D- isomer, and given that the optical rotation of the D- isomer is + 13.5°, the L- isomer's optical rotation will have the same magnitude but an opposite sign, resulting in -13.5°.

Thus, the rotation of the racemic mixture will be equal to 0°.

- This occurs because a racemic mixture contains equal proportions of both enantiomers.

Convert HCl and H2O to moles.

36.0 g of HCl = 0.987 moles HCl

98.0 g of H2O = 5.44 moles H2O

Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.

For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.

Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O

Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.

11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O

11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.

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!!!