Helpppp!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

1 answer:

castortr0y [4]3 years ago

6 0

Answer:

mk, ill try. also im listing them left to right

Explanation:

first row: carbon, oxygen, chlorine, nitrogen

second row: chlorine, glucose, sodium chloride, cobalt

third: hydrogen chloride (i dont know the others and dont want to get them wrong but hope this helped in some sort of way)

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if the theoretical yield of a reaction is 26.0 grams and you actually recovered 22.0 grams what is the precent yield

netineya [11]

Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

∴ if theoretical yield is 26 g, but only 22.0 is recovered from the reaction,
then Percentage Yield = (22 g ÷ 26 g) × 100
= 84.6 %

8 0

3 years ago

A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution

Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions× $\frac{1(NH_{4})_{2}SO_{4}}{3Ions}$ = 0,262 moles of (NH₄)₂SO₄