Magnesium has an atomic mass of 24.3. there are two isotopes of magnesium - one contains 12 neutrons and the other contains 13 n

The ether layer will consist of aniline and toluene following the extraction process. Explanation: Solvent extraction separates compounds based on their solubilities in two distinct immiscible liquids; typically, one being water and the other, an organic solvent like ether in this instance. PhCO₂H (benzoic acid) interacts with NaOH, resulting in a highly soluble benzoate ion in water yet poorly soluble in ether, thus it remains in the aqueous layer. Conversely, PhNH₂ (aniline) dissolves well in organic solvents such as ether while having minimal or no solubility in water; PhCH₃ (toluene) possesses solubility in ether whilst being insoluble in water. Hence, post-extraction, the ether layer encompasses aniline and toluene.

Answer:

b) 0.47

Explanation:

Molar mass of C5H12 = 72.15g/mol

⇒ moles of C5H12 = (10.0)*(mol/72.15)=0.1386 moles of C5H12

Molar mass of C6H14 = 86.18g/mol

⇒ moles of C6H14 = (20.0)*(mol/86.18)=0.232 moles

Molar mass of C6H6 = 78.11g/mol

⇒ moles of C6H6 = (10.0)*(mol/78.11)=0.128 moles of C6H6

Therefore, XC6H14=(0.232)/(0.1386+0.232+0.128)=0.465≅0.47

Answer:

The mass percentage of GaBr₃ in the solid mixture is 30.2 %.

Explanation:

GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)

To find the molar mass, we have: MW GaBr₃ = 309.4 g/mol and MW AgBr = 187.8 g/mol.

Using these values, we note that 187.8 g of AgBr corresponds to 1 mol.

If 0.368 g of AgBr yields x moles, we calculate x to be 2.0 x 10⁻³ mol AgBr.

Based on the stoichiometry, 1 mol of GaBr₃ produces 3 mol of AgBr, giving us:

y ___________ 2.0 x 10⁻³ mol AgBr

Thus, y equals 6.7 x 10⁻⁴ mol GaBr₃.

Additionally, 1 mol of GaBr₃ has a mass of 309.4 g.

Consequently, 6.7 x 10⁻⁴ mol of GaBr₃ corresponds to w grams of GaBr₃, yielding w = 0.206 g GaBr₃.

Relating this to the total mass:

0.6813 g _____ 100%

0.206 g _____ z

Therefore, z is calculated as 30.2 %.

The chalk production's efficiency is noted to be 82 %

To determine this efficiency, we utilize the formula,

% Efficiency

Finding the actual yield based on % efficiency and the theoretical yield:

82 %

Actual yield =

Each box contains 145 grams of chalk

Total boxes =

The chalk box company aims to produce 3000 boxes. However, with 82 % efficiency, they can only produce 2262 boxes, thus falling short of their target.