Answer:
% recovery
MP range of product
mass of product
Explanation:
Liquid–liquid extraction (LLE) is a process of transferring one (or more) solute(s) which are present in a feed solution to another immiscible liquid (solvent). The other solvent that becomes enriched in the target solute(s) is called extract. The original feed solution that is depleted in solute(s) is subsequently referred to as the raffinate.
This method is used to purify compounds and separate mixtures of compounds. This is very important when we want to isolate a product from a reaction mixture.
The percent recovery is the amount of solute that is transferred to the extract. This is the most important data to be recorded in an LLE experiment.
The melting point range necessarily helps us to identify the product and the mass of solid tells us the quantity of the solid obtained after extraction.
Answer:
2 M
Explanation:
mole weight of CaBr2 = 40 + 2 * 79.9 = 199.8 gm
20 gm is then 20/199.8 =.1 mole
.1 mole / .50 liter = 2 M
Find the moles of BaSO4 first. Then since we know it's a one to one ratio from barium chloride to barium sulfate we can just solve for liters.
<span>First you need to find the moles BaSO4 , and the you will require to find barium sulfate in liters.
</span>12.00gBaSO4 / 233.31 grams per mole
=.05141moles
Molarity=moles/liters
Hence,
Liters=.05141moles/.6Molarity
=.85 liters
From the combustion of octane, the formaldehyde will be formed as this equation:
C8H18 + O2 → CH2O + H2O this is the original equation but it is not a balanced equation, so let's start to balance it:
the equation to be balanced so the number of atoms on the right side of the equation sholud be equal with the number of atoms on the lef side.
-we have 8 C atoms on left side and 1 atom on the right side so we will try putting 8 CH2O on the right side instead of CH2O
C8H18 + O2 → 8 CH2O + H2O
we have 2 O atoms on the left side and 9 atoms on the right side so we will try first to put 9 O2 instead of O2 on the left side and put 2H2O on the right side and put 16 CH2O instead of 8 CH2O to make the atoms of O are equal on both sides = 18 atoms
C8H18 + 9 O2 → 16 CH2O + 2H2O
put now we have 8 atom C on the left side and 16 atom on the right side so, we will put 2 C8H18 instead of C8H18 now we get this equation:
2C8H18 + 9O2 →16 CH2O + 2H2O
-now we have 36 of H atoms on both sides.
- and 16 of C atoms on both sides.
- and 18 of O atoms on both sides.
now all the number of atoms of O & C & H are equal on both sides
∴ 2C8H18 + 9O2 → 16 CH2O + 2 H2O
is the final balanced equation for the formation of formaldehayde