Explanation:
It is known that the molecular weight of is 111 g/mol. This means that 1 mole of contains 111 g .
1 g =
As we know that the density of water is 1 g/cc (as 1 ml = 1 cc).
So, 100 ml water = 100 g water. Therefore, in 100 g of water present will be calculated as follows.
mol
So, in 1000 g water the amount of present will be calculated as follows.
= 0.09 mol
Hence, the molality of is 0.09 mol.
According to Raoult's law,
where, = boiling point constant
For pure 1 kg water, = 0.52 K.kg/mol
m = molality of solution
Therefore, putting the given values into the above formula as follows.
=
= 0.0468 K
Therefore, the boiling point will raise by 0.0468 K.
You would use a Graduated Cylinder
Answer:
Explanation:
Given that;
diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m
length (l) = 10 cm = 0.1 m
porosity = 50%
height (h) = 30 cm = 0.3 m
time (t) = 5 s
volume (v) = 60 cm³ = 60 × 10⁻⁶ m³
Q (flow rate) =
Q =
Q =
From constant head method, we use the relation K = to determine the hydraulic conductivity ; we have:
Seepage velocity
where; velocity =
=
=
Protons and neutrons, just like in the nucleus of every other element.