Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.
Explanation:
Conversion of degree Celsius to Kelvin :
K=^oC+273
Conversion of degree Celsius to degrees Fahrenheit :
^oF=(\frac{9}{5}\times ^oC)+32
By using these two conversion factors, we get the three temperature readings all mean the same thing.
For option A :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option B :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option C :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
For option D :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
From the given options, only option (D) is correct.
Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
Hope this helps!
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Hi Sydney!
I can't draw in this question, but there is a picture showing this phase for you to follow when you draw it.
Hope This Helps :)
Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
