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irga5000 [103]
2 years ago
8

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.500 atm? (∆Hvap = 28.5

kJ/mol)
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
7 0

For a normal boiling point of a liquid is 282 °C, the temperature is mathematically given as

T2=181.55°C\

<h3> What temperature (in °C) would the vapor pressure be 0.500 atm? </h3>

Generally, the equation for the gas  is mathematically given as

ln(p1/p2)=dHvap/R(1/T2-1/T1)

Therefore

ln(1/0.26)=23500/8.214(1/T2-1/555)

T2=181.55^C

In conclusion

T2=181.55°C

Read more about Temperature

brainly.com/question/13439286

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Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

<u>Data:</u>

M1 = 6.01 M stock solution concentration

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V1 = ? stock solution volume

M1 * V1 = M2 * V2

V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL

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3 years ago
Why does temperature change always move from regions of hot to regions of cold?
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A binary compound created by reaction of bismuth and an unknown element E contains 52.07% Bi and 47.93% E by mass. If the formul
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Answer:

Atomic mass of E is 128.24

Explanation:

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Percentage composition of bismuth = 52.07%

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Mass Bismuth in the compound is (2×208.9804) = 417.96 g

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To calculate the atomic mass of E

52.07% = 417.96 g

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            = (47.93 × 417.96 ) ÷ 52.07 %

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Therefore; E = 384.729 ÷ 3

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