We know the volume of one mole of gas at 273 K and 760 Torr is 22.4 L. Using
(PV)/T = constant
We can calculate the volume of the gasses at the given conditions:
(P₁V₁)/T₁ = (P₂V₂)/T₂
(760 * 22.4) / 273 = (288 V₂) / 308.2
V₂ = 66.7 L
Mass of He: 4
Mass of Ne: 20
Fraction of Ne: x
Fraction of He: 1 - x
avg density = (∑(component fraction × component mass))/volume
0.2460 = (20x + 4(1 - x))/ 66.7
x = 0.775
Answer:
The heat required to change 25.0 g of water from solid ice to liquid water at 0°C is 8350 J
Explanation:
The parameters given are
The temperature of the solid water = 0°C
The heat of fusion, = 334 J/g
The heat of vaporization, = 2260 J/g
Mass of the solid water = 25.0 g
We note that the heat required to change a solid to a liquid is the heat of fusion, from which we have the formula for heat fusion is given as follows;
ΔH = m ×
Therefore, we have;
ΔH = 25 g × 334 J/g = 8350 J
Which gives the heat required to change 25.0 g of water from solid ice to liquid water at 0°C as 8350 J.
Ions are atoms or groups of atoms that have either gained or lost electrons
From the ideal gas law
pv=nRT , n is therefore PV/RT
R is the
R is gas constant =62.364 torr/mol/k
P=500torr
V=4.00l
T=500+273=773k
n={(500 torr x 4.00l)/(62.364 x773k)}=0.041moles
the number of molecules=moles x avorgadro costant that is 6.022x10^23)
6.022 x 10^23) x0.041=2.469 x10^22molecules
