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3 years ago

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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3 years ago

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g of CO2 and

vladimir2022 [97]

Answer:

$C_{7} H_{5}N_{3}O_{6}$

Explanation:

First reaction gives you the number of moles or the mass from Carbon and hydrogen

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$0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC$

$0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC\\$

Analogously for hydrogen:

0.0310g $H_{2}O$ have 0.0034gH or 0.0034mol of H

In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.

$0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N$

now

$\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction$

this is equivalet to 0.002mol of N

with this information you can find the mass of oxygen by matter conservation.

$gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO$

this is equivalent to 0.004molO

finally you divide all moles obtained between the smaller number of mole (this is mol of H)

$C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}$

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3 years ago

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3 years ago

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3 years ago

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Anarel [89]

Answer:

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Explanation:

Wolf-Kishner reaction is a type of reduction reaction in which aldehydes and ketones are reduced to their corresponding alkane in the presence of a base.

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3 years ago