5 deg C = 41 deg F.
The freezing point of water is 0°C or 273K in correct SI units.
increasing the temperature increases the intermolecular spaces and decreases the intermolecular forces,thus increasing ideality.... so at high temperature of 327c sulphurdioxide is ideal as compared to 273k
Assuming the K means Kelvin - which in itself is a measure of temperature.31.73 degrees Fahrenheit-0.15 degrees Celsius
146c,351k, 133f, 95f, 25c, 58f, 273k
its 0C and 273K
No. There are no negative kelvin temperatures.
273K = 0oC Hence 98K = 273K - XoC XoC = 98K - 273K = -175oC
The lowest possible temperature is -273.15 C which equated to '0'K. Therefore, temperature -273K cannot exist. No -ve K temperatures exist.
Nope … you have it backward. 0 Centigrade equals (plus) 273 degrees Kelvin.
-273k
273K = -0.15ºCUse this equation to convert Kelvin to degrees Celsius/Centigrade: [°C] = [K] - 273.15
-273k
The freezing point of water is 0°C or 273K in correct SI units.
Freezes.
273K
At 273K and 1 ATM, most gases typically exist in the gaseous state. However, the specific gas present will depend on the composition of the system.
1C= 273k