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swat32
3 years ago
14

Perform the following conversions:

Chemistry
1 answer:
ololo11 [35]3 years ago
8 0

Answer:

a) 20^0C and 293K

b) 437K and 327.2^0F

c) -273^0C and 459.44^0F

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like ^0C , ^0Fand K  

These units of temperature are inter convertible.

t^0C=(t+273)K

t^oC=\frac{5}{9}\times (t^oF-32)

K-273=\frac{5}{9}\times (^oF-32)

a)  68°F (a pleasant spring day) to °C and K.

Converting this unit of temperature into ^0C and K by using conversion factor:

t^oC=\frac{5}{9}\times (68^oF-32)

t=20^0C

20^0C=(20+273)K=293K

b) 164°C (the boiling point of methane, the main component of natural gas) to K and °F

Conversion from degree Celsius to Kelvins  and Fahrenheit

164C=\frac{5}{9}\times (t^oF-32)

t^0F=327.2

164°C=(164+273)K=437 K

c) 0K (absolute zero, theoretically the coldest possible temperature) to °C and °F.

K-273=\frac{5}{9}\times (t^oF-32)

0-273=\frac{5}{9}\times (t^oF-32)

t=-459.4^0F

t^K=(0-273)^0C=-273^0C

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