I hope that you can understand this!!! Lol
The thing that you have to pull back to release with, that would be considered a third class lever.
I hope this helps. :)
Answer:
[CaCl₂·2H₂O] = 1.43 m
Explanation:
Molality is mol of solute / kg of solvent.
Mass of solvent = 40 g
Let's convert g to kg → 40 g / 1000 = 0.04 kg
Let's determine the moles of solute (mass / molar mass)
8.43 g / 146.98 g/mol = 0.057 mol
Molality = 0.057 mol / 0.04 kg → 1.43
The answer would be 2.0 x 10^-1
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
