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soldier1979 [14.2K]
3 years ago
11

Which common material is an example of a polyamide?

Chemistry
1 answer:
Vlad1618 [11]3 years ago
5 0

Nylon 6,6 is a common example of a polyamide.

<em>Polyamides</em> are polymers that contain <em>repeating amide (-CO-NH-) linkages</em>.

The structure of Nylon 6,6 is  

[-NH-(CH_2)_6-<u>NH-CO</u>-(CH_2)_4-CO-]_<em>n</em>

where <em>n</em> is a large number.

The numbers in the name showow that there are six carbon atoms on either side of an amide linkage.

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What is the molar mass of Na2CO3? 60.0 g/mol 106.0 g/mol 118.0 g/mol 141.0 g/mol
Dmitry [639]

Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.

M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).

M(Na₂CO₃) = 2 · 23 + 12 + 3 · 16 · g/mol.

M(Na₂CO₃) = 46 + 12 + 48 · g/mol.

M(Na₂CO₃) = 106 g/mol; molar mass of sodium carbonate.

Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.

8 0
3 years ago
What percentage of the filtrate's water that enters bowman's capsule is reabsorbed into the blood? select one:
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It asks for the percentage but I’m not sure how to find it when the answer is 67.1%
Vlada [557]
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4 0
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If the farmer had weighed out 32.0 pounds of chicken eggs rather than kilograms what mass of quail eggs would he need to weigh o
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8 0
3 years ago
A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.
n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s  at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) =  (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

 ⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1  = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

Ea = (R*ln (k2/k1)) / ((1/T1)- (1/T2))

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1   - 1/ T3   = 1/400   - 1 /425    = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

4 0
3 years ago
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