Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
The property to be used is sieving. you get a bowl and a sieve when you pour the sand and salt,the salt being the small particle sieve to the bowl while the sand remain on the sieve as residue
<span>The slim exit of the
column is first persisted with glass wool or a permeable plate in order to sustain
the column packing element and keep it from getting out of the tube. Then the
adsorbent solid, which is usually a silica, is firmly packed into the glass
tube to make the separating column. The packing of the non-moving phase into
the glass column must be done with precaution to create an even distribution of
material. An even distribution of adsorbent material is very important to lessen
the existence of air bubbles and/or channels inside the column. To finish
preparing the column, the solvent to be used as the mobile phase is delivered
through the dry column. Then the column is said to be "wetted" and
the column must stay wet throughout the entire procedure. Once the column is properly
prepared, the sample to be separated is placed at the top of the wet column.</span>
The remaining 3 % are nanomaterials made from e.g. aluminium oxide, barium titanate, titanium dioxide, cerium oxide and zinc oxide. Carbon nanotubes, graphene and fullerenes have annual production amounts in the hundred tonnes range. Nanosilver is estimated to be produced in about 20 tonnes per year.
I looked this up but hope it helps
