The only two elements that are liquid at room temperature are Bromine (Br) and Mercury (Hg).
Answer: It would be the boiling point.
Explanation:
Because it is not a liquid nor a gas, that would mark the last two out. I'll have to guess that it is the boiling point of water.
A sample of liquid benzene at 80°C and a sample of gaseous benzene at 80°C will have the same average kinetic energy of benzene molecules.
Let us recall that the average kinetic energy of molecules depends on the molar mass of the substance as well as the temperature of the substance. We have that;
average kinetic energy= ![\frac{3RT}{2NA}](https://tex.z-dn.net/?f=%5Cfrac%7B3RT%7D%7B2NA%7D)
R = gas constant
NA =Number of molecules
M = Temperature in Kelvin
We can see that the average kinetic energy depends on temperature therefore, a sample of liquid benzene at 80°C and a sample of gaseous benzene at 80°C will have the same average kinetic energy of benzene molecules.
Learn more: brainly.com/question/6284546
Answer:
P = 5.868 atm
Explanation:
- Pt = (PN2O)(XN2O) + (PO2)(XO2)
assume T = 25°C = 298.15 K
∴ PN2O = 1 atm
∴ XN2O = nN2O/nt
∴ n = PV/RT....ideal gas
⇒ nN2O = ((1atm)(1L))/((0.082atm.L/K.mol)(298.15K)) = 0.041 mol N2O
∴ PO2 = 6 atm
∴ XO2 = nO2/nt
∴ nO2 = ((6atm)(6L))/((0.082atm.L/K.mol)(298.15K)) = 1.472 mol O2
⇒ nt = nN2O + nO2 = 0.041 + 1.472 = 1.513 mol
⇒ XN2O = 0.041/1.513 = 0.03
⇒ XO2 = 1.472/1.513 = 0.973
⇒ Pt = ((1atm)(0.03)) + ((6atm)(0.973))
⇒ Pt = 5.868 atm
Explanation:
It is known that density is the amount of mass present in liter of solution or substance.
Mathematically, Density = ![\frac{mass}{volume}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D)
It is given that volume is 3.25 L and mass is
. Hence, calculate the density of glycerol as follows.
Density = ![\frac{mass}{volume}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D)
= ![\frac{4.10 \times 10^{3} g}{3.25 L}](https://tex.z-dn.net/?f=%5Cfrac%7B4.10%20%5Ctimes%2010%5E%7B3%7D%20g%7D%7B3.25%20L%7D)
= ![1.26 \times 10^{3} g/L](https://tex.z-dn.net/?f=1.26%20%5Ctimes%2010%5E%7B3%7D%20g%2FL)
As, 1 L = 1000
.
So, ![1.26 \times 10^{3} g/L \times \frac{1000 cm^{3}}{1 L}](https://tex.z-dn.net/?f=1.26%20%5Ctimes%2010%5E%7B3%7D%20g%2FL%20%5Ctimes%20%5Cfrac%7B1000%20cm%5E%7B3%7D%7D%7B1%20L%7D)
=
Thus, we can conclude that the density of glycerol is
.