If 3.11 mol of an ideal gas has a pressure of 2.91 atm and a volume of 78.13 L, what is the temperature of the sample in degrees Celsius?

617.35 °C

Explanation:

Using the ideal gas law,

PV = nRT

P = Pressure in (atm or pascal)

V = Volume in (litres or cubic metres)

n = number of moles.

R = Gas constant in (0.0821L .atm/mol.K or 8.314J/mol/K)

T = Temperature in degree Kelvin or Celsius.

PV = nRT

P = 2.91atm, V= 78.13L , n= 3.11

R = 0.0821L .atm/mol.K

T = ?

T = 2.91 x 78.13 / 3.11 x 0.0821 = 227.3583/ 0.255331 = 890.5 K

Converting from Kelvin to Celsius, using: °C = K - 273.15

= 890.5 - 273.15

°C = 617.35

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

2b2t

explanation:

2b2t