Question

The given table shows

the freezing points in °F of

different gases at sea level.

Convert each of these into °

C to the nearest integral

value using the relation and

complete the table, C = 5/9 (°F –

32)

where, °C stands for degree

Celsius,and °F stands for

degree Fahrenheit.

Gas Freezing point at

Sea level (°F)

Freezing point at sea

level (°C)

Hydrogen -435 Krypton -

251 Oxygen -369

Helium -458 Argon -309​

Answer 1

Answer:

$c = \frac{(5)(f - 32)}{9} \\ c = \frac{5f - 160}{9}$

(i) Hydrogen - 435°F

$c = \frac{5f - 160}{9} \\ c = \frac{(5)(435) - 160}{9} \\ c = \frac{2175 - 160}{9} \\ c = \frac{2005}{9} \\ c = 223 \frac{7}{9}$

(ii) Krypton - 251

$c = \frac{5f - 160}{9} \\ c = \frac{(5)(251) - 160}{9} \\c = \frac{1255 - 160}{9} \\ c = \frac{1095}{9} \\ c = 121 \frac{6}{9}$

Similarly, other questions can be solved.

Hope this helps.

Thanks.

Answer 2

Step-by-step explanation:

For hydrogen;

F∴C=−4350F=95(−435−32)=95(−467)=−92335=−259.44∘C

For Krypton;

FC=−2510F=95(−251−32)=95(−283)=9−1415=−157.22∘C

For Oxygen;

F=−3690F

C=95(−369−32)=95(−401)=9−2005=−222.77∘C

For Helium;

F=−4580F

C=95(−458−32)=95(−490)=9−2450=−272.22∘C

For Argon;

F=−3090F

C=95(−309−32)=95(−341)=9−1705=−189.44∘C