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dem82 [27]
2 years ago
7

Describe the structure of ammonium lauryl sulfate. Refer to the given diagram. Your answer should include the type of bonding, t

he elements contained, and the size and shape of the molecule. Write a short paragraph.

Chemistry
1 answer:
Mariana [72]2 years ago
3 0

Explanation:

Ammonium lauryl sulfate (ALS) is the common name for ammonium dodecyl sulfate (CH3(CH2)10CH2OSO3NH4). The anion consists of a nonpolar hydrocarbon chain and a polar sulfate end group. The combination of nonpolar and polar groups confers surfactant properties to the anion: it facilitates the dissolution of both polar and non-polar materials. ALS is classified as a sulfate ester. It is found primarily in shampoos and body wash as a foaming agent. Lauryl sulfates are very high-foam surfactants that disrupt the surface tension of water in part by forming micelles at the surface-air interface.

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
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<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

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rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
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