Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg
Light because it is an electromagnetic wave and they can travel through the vacuum in outer space.
B, because gold is the most malleable substance of all.
The energy delivered by the laser in 1 second is

In order to find how many photons correspond to this energy, we must calculate the energy of a single photon.
Calling h the Planck constant, c the speed of light and

the wavelength of the light, the energy of a single photon is given by

So, the number of photons emitted by the laser in 1 second is equal to the total energy delivered by the laser divided by the energy of a single photon:

photons