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.
0.045 mol L 0.5. Mol.
no kelvin is used to measure temperature it ranges from 273K to 373K
You can use the Ideal Gas Law or the Combined Gas Law. PV/T = PV/T ---- left are usually labeled (1) and right (2) to show different conditions left P = unknown, V = 19.6L, T = (273+27)K right P= 760 torr or 1 amt, V = 22.4 L, T = 273K --- all are standard P(19.6L)/300K = 1atm(22.4L)/273K --- solve for P P = (1 atm)(22.4L)(300K)/ [(273K)(19.6L)] do the math --- get the answer.
To get from Celsius to Kelvin, +273 And obviously, -273 to get from Kelvin to Celsius E.g. -273oC = 0K (Absolute zero) 0oC = 273K (Freezing temp. of water) 25oC = 298K (Standard temp.)
-273k
273K = -0.15ºCUse this equation to convert Kelvin to degrees Celsius/Centigrade: [°C] = [K] - 273.15
273K = 0oC Hence 98K = 273K - XoC XoC = 98K - 273K = -175oC
-273k
The freezing point of water is 0°C or 273K in correct SI units.
Freezes.
The lowest possible temperature is -273.15 C which equated to '0'K. Therefore, temperature -273K cannot exist. No -ve K temperatures exist.
273K
B. 273K. 5 degrees Celsius is equal to 278K in the Kelvin scale, but the closest value among the given options is 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
No. There are no negative kelvin temperatures.