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Liula [17]
3 years ago
11

1. Phosphorous reacts with bromine to form phosphorous tribromide. If 35.0 grams of bromine

Chemistry
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

70.6 %

Explanation:

First step, we define the reaction:

2P + 3Br₂  →  2PBr₃

We determine the moles of reactant:

35 g . 1mol / 159.8 g = 0.219 moles

We assume, the P is in excess, so the bromine is the limiting reagent.

3 moles of Br₂ can produce 2 moles of phophorous tribromide

Then, 0.219 moles may produce (0.219 . 2) /3 = 0.146 moles of PBr₃

We convert moles to mass:

0.146 mol .  270.67 g /mol = 39.5 g

That's the 100 % yield reaction, also called theoretical yield. The way to determine the % yield is:

(Yield produced / Thoeretical yield) . 100

(27.9 / 39.5) . 100 = 70.6 %

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Answer: V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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where,

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T_2 = final temperature of gas = 283K

Now put all the given values in the above equation, we get the final volume of gas.

V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

V_2=20.4ml

Thus the correct numerical setup for calculating the new volume is \frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

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