We are given that the balanced chemical reaction is:
cacl2⋅2h2o(aq) +
k2c2o4⋅h2o(aq) --->
cac2o4⋅h2o(s) +
2kcl(aq) + 2h2o(l)
We known that
the product was oven dried, therefore the mass of 0.333 g pertains only to that
of the substance cac2o4⋅h2o(s). So what we will do first is to convert this
into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is
molar mass of cac2o4 plus the
molar mass of h2o.
molar mass cac2o4⋅h2o(s) = 128.10
+ 18 = 146.10 g /mole
moles cac2o4⋅h2o(s) =
0.333 / 146.10 = 2.28 x 10^-3 moles
Looking at
the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is
1:1, therefore:
moles k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles
Converting
this to mass:
mass k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles (184.24 g /mol) = 0.419931006 g
Therefore:
The mass of k2c2o4⋅<span>h2o(aq) in
the salt mixture is about 0.420 g</span>
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?
volume of water added = final volume -initial volume
= 120-10
=110ml
Answer:
The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass
Explanation:
2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.
We know that density is equal to ratio of mass to volume.
Here density of solution is 1.01 g/mL.
So mass of 1000 mL solution = (
) g = 1010 g
molar mass of NaCl = 58.44 g/mol
So mass of 2.23 moles of NaCl = (
) g = 130.3 g
% by mass is ratio of mass of solute to mass of solution and then multiplied by 100.
Here solute is NaCl.
So % by mass of 2.23 M aqueous solution of NaCl = 
% = 12.90%
Answer:
B. n-octyl alcohol and 1-octene
Explanation:
Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. The principle is that different compounds in the sample mixture travel at different rates due to the differences in interactions with stationary phase and due to the differences in solubility in the solvent. The principal chemical property for separation using this technique is molecular polarity
You can intuit than hexadecane and octadecane don't have big polarity differences, also chlorobenzene and bromobenzene haven't.
An alcohol as n-octyl alcohol has different polarity than an alkene as 1-octene.
Thus, using thin layer chromatography is most easy to separate:
<em>B. n-octyl alcohol and 1-octene
</em>
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I hope it helps!
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Guess and check, test, trial and error, completion.
