Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers: (a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O (b) Saran; 24.8% C, 2.0% H, 73.1% Cl (c) polyethylene; 86% C, 14% H (d) polystyrene; 92.3% C, 7.7% H (e) Orlon; 67.9% C, 5.70% H, 26.4% N
Answers
Answer:
(a)
(b)
(c)
(d)
(e)
Explanation:
Hello,
(a) For the lucite, one computes the moles of C, H and O that are present:
Now, dividing each moles by the smallest moles (oxygen's moles), one obtains:
Finally, we look for the smallest whole number subscript by multiplying by 2, so the empirical formula turns out into:
(b) For the Saran, one computes the moles of C, H and Cl that are present:
Now, dividing each moles by the smallest moles (hydrogen's moles), one obtains:
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
(c) For the polyethylene, one computes the moles of C and H that are present:
Now, dividing each moles by the smallest moles (carbon's moles), one obtains:
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
(d) For the polystyrene, one computes the moles of C and H that are present:
Now, dividing each moles by the smallest moles (either carbon's or hydrogen's moles), one obtains:
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
(e) For the orlon, one computes the moles of C, H and N that are present:
Now, dividing each moles by the smallest moles (nitrogen's moles), one obtains:
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
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The amount of I − 3 ( aq ) in a solution can be determined by titration with a solution containing a known concentration of S 2 O 2 − 3 ( aq ) (thiosulfate ion). The determination is based on the net ionic equation 2 S 2 O 2 − 3 ( aq ) + I − 3 ( aq ) ⟶ S 4 O 2 − 6 ( aq ) + 3 I − ( aq ) Given that it requires 29.4 mL of 0.380 M Na 2 S 2 O 3 ( aq ) to titrate a 30.0 mL sample of I − 3 ( aq ) , calculate the molarity of I − 3 ( aq ) in the solution.
An aerosol is a colloidal _____ suspended in a _____. A.liquid B.solid C.gas
You have been injured in the laboratory (cut,burn,etc) first you should
Answers
Answer:
Esterification reaction
Explanation:
When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).
In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.
Answers
Answers
Answer:
1. Bronsted—Lowry acid
2. Bronsted—Lowry Base
3. Lower the pka
4. Lewis acids
Explanation:
Answers
Answer:pH = 2.96
Explanation:
C5H5N + HBr --------------> C5H5N+ + Br-
millimoles of pyridine = 80 x 0.3184 =25.472mM
25.472 millimoles of HBr must be added to reach equivalence point.
25.472 = V x 0.5397
V =25.472/0.5397= 47.197 mL HBr
total volume = 80 + 47.197= 127.196 mL
Concentration of [C5H5N+] = no of moles / volume=
25.472/ 127.196= 0.20M
so,
pOH = 1/2 [pKw + pKa + log C]
pKb = 8.77
pOH = 1/2 [14 + 8.77 + log 0.20]
pOH = 11.0355
pH = 14 - 11.0355
pH = 2.96
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.
Answers
Answer:
the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
Explanation:
Given that;
The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K
Initial volume of the container, V1 = 537 mL
Initial vapor pressure, P1 = 68.0 mmHg
Final volume of the container, V2 = 338 mL
Let us say that the final vapor pressure = P2
From Boyle's law,
P2V2 = P1V1
P2 * 338 = 68.0 * 537
338P2 = 36516
P2 = 36516 / 338
P2 = 108.03 mmHg
Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.
Hence,
b. Some of the vapor initially present will condense.
e. Liquid octane will be present.
True/False
Answers
Answer:
True
Explanation:
Liquid hexane burns in oxygen gas to form carbon dioxide gas and water vapor.TRUE.
This is the complete combustion reaction of hexane, which proceeds according to the following equation.
C₆H₁₄(l) + 9.5 O₂(g) → 6 CO₂(g) + 7 H₂O(g)
If the combustion were incomplete, instead of carbon dioxide, the product would be carbon monoxide or carbon.