18:25 ll 15
You are given an unknown Substance and
told that it could be one of the three substances.
L. Mor N. You developed the two sets of
chromatograms below. From the chromatograms,
deduce the identity of X
台
solvent -
front
starting -
line
XLMN
developed
with water
XL M N
developed
with ethanol
A It could be either Lor M.
B It could be either Mor N.
C It must be La
DIt must be N.
Problem-solving strategy:
If two substances are identical, they will
have the same height on a chromatogram
regardless of the solvent used.
3. Dyes that are allowed in loods are given E
numbers. Four such E numbers commonly
used in soups are E120, E122, E124 and E162.
The chromatogram below was obtained from
three tomato soups X, Y and Z. Soup 2 was a
new soup on the market. A chemist was testing
it to ensure that all of the dyes used in making
the soup were legal.
Х
Y
Z
E120 122 E124 E162
a) Explain wity water can be used as a solvent.
b) Why was the pencil line, on which the spots
were placed, drawn above the level of the
solvent?
c) Which soup contains all lour E numbers?
d) Are the E numbers in soup 2 legp? Explain
your awer.
IM​

Ireally don’t know this

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³