False, although the air is moving in an equilibrium us normal humans can not see the air actually moving. So it is false.
What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have
or
where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.
Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R.
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Thus, we have
P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L
Now, applying these converted values into the gas law, we have
Given that the mass of the sample is 0.430 g, we have
Thus, the gas sample has a molar mass of 43.9 g/mol.
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Explanation:
the molar mass of a compound can be caucaleted by adding the standar atomic masses.