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laila [671]
3 years ago
7

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical for

mulas. 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%
Chemistry
2 answers:
sp2606 [1]3 years ago
6 0

Answer:

(a) C5H8O2

(b) C2H2Cl2

(c) CH2

(d) CH

Explanation:

We need to find the proportion of the atoms in whole numbers. Given the percentages we can calculate the number of moles and find their proportions.

Assume 100 g and given the atomic weights the moles are calculated.

(a) C = 59.9/ 12.01 = 4.98  ≈ 5.00

    H = 8.06/1.007 = 8.00

    O = 32/15.999  = 2.00

C5H8O2

(b) C= 24.8/12.01 = 2.06≈ 2.00

    H = 2.0/1.007 = 1.99 ≈ 2.00

   Cl = 73.1/ 35.453 = 2.06 ≈ 2.00

C2H2Cl2

(c) C = 86/12.01 = 7.16  

    H= 14/1.007 = 13.90

7.16:13.90 ≈ 1:2

CH2

(d) C = 92.30/12.01 = 7.68

    H = 7.7 / 1.007 =   7.65

7.68:7.65 ≈ 1:1

CH

svet-max [94.6K]3 years ago
5 0

In letter d that's a missing information: it's 7.7% H.

Answer:

a. C₅H₈O₂

b. CHCl

c. CH₂

d. CH

Explanation:

Let's consider a sample that has 100 g of mass, so by that, we can calculate the mass of each component. After that, the number of moles can be find dividing the mass by the molar mass. The molar mass of the elements given are: C = 12 g/mol, O = 16 g/mol, H = 1 g/mol, and Cl = 35.5 g/mol. The empiric formula is the simplest one, so, all the number moles must be divided by the small one. If it's not divisible, then the empiric formula is already done.

a.

mC = 0.599*100 = 59.9 g

mH = 0.0806*100 = 8.06 g

mO = 0.32*100 = 32 g

nC = 59.9/12 = 5 mol

nH = 8.06/1 = 8 mol

nO = 32/16 = 2 mol

Because 5 is not divided by 2, the empiric formula is C₅H₈O₂.

b.

mC = 0.248*100 = 24.8 g

mH = 0.02*100 = 2 g

mCl = 0.731*100 = 73.1 g

nC = 24.8/12 = 2 mol

nH = 2/1 = 2 mol

nCl = 73.1/35.5 = 2 mol

So, all must be divided by 2, and the empiric formula will be CHCl.

c.

mC = 0.86*100 = 86 g

mH = 0.14*100 = 14 g

nC = 86/12 = 7 mol

nH = 14/1 = 14 mol

Dividing all by 7, the empiric formula is CH₂.

d.

mC = 0.923 *100 = 92.3 g

mH = 0.077*100 = 7.7 g

nC = 92.3/12 = 7.7

nH = 7.7/1 = 7.7

Dividing all by 7.7, the empirical formula is CH.

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To calculate the number of moles, simply take the ratio of mass and molar mass. The molar mass of Iron (Fe) is equal to 55.85 g/mol.Therefore the total number of moles is:
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7 0
3 years ago
What volume (in L) of a 1.25 M
sleet_krkn [62]

Answer:

V1= 0.305L

Explanation:

To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2

Since the question wants the volume in litres, convert 455 mL to L

455/ 1000

= 0.455 L

Now make the substitution

1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

V1=

\frac{0.838 \times 0.455}{1.25 }  = 0.305

4 0
4 years ago
How many moles of hydrogen are produced when 6.28 mil of oxygen form
sergejj [24]
If im correct i think the answer is 12.6 mol.
6 0
3 years ago
Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V
malfutka [58]

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

4 0
3 years ago
Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i
Goshia [24]

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 3.073\times 10^4kJ/L respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

\Delta H^o_{comb}=-3535kJ/mol

Fuel value = \frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}

Molar mass of pentane = 72 g/mol

Fuel value = \frac{3535kJ/mol}{72g/mol}

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = 3.073\times 10^4kJ/L

Thus, the fuel density of pentane is 3.073\times 10^4kJ/L

4 0
4 years ago
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