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MArishka [77]
2 years ago
10

The DEA has designated five chemicals often found in spice as _____ controlled substances, making it illegal to sell, buy, or po

ssess these chemicals.
Chemistry
1 answer:
Mrrafil [7]2 years ago
4 0

Answer:

I'm pretty sure this doesn't really connect with chemistry much but the answer should be <u>Schedule 1 </u> as stated by the Drug enforcement agency.

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Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B
liraira [26]

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

6 0
2 years ago
What factors affect the rate of dissolving a solid in a liquid? Explain...
tester [92]

stirring, temperature, and particle size

4 0
3 years ago
If the density of pure water is 0.9922 g/mL at 40 ºC, calculate its theoretical molarity at that temperature. Report to 4 sig fi
OleMash [197]
Answer is: theoretical molarity of water is 55.1222 mol/L.<span>
d(H</span>₂O) = 0.9922 g/mL.
M(H₂O) = 2 · Ar(H) + Ar(O) · g/mol.
M(H₂O) = 2 + 16 · g/mol = 18 g/mol.
c(H₂O) = d(H₂O) ÷ M(H₂O).
c(H₂O) = 0.9922 g/mL ÷ 18 g/mol.
c(H₂O) = 0.0551 mol/mL.
c(H₂O) = 0.0551 mol/mL · 1000 mL/L = 55.1222 mol/L.
3 0
3 years ago
You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that
Alecsey [184]

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

                                     = 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

                        = 0.005 g

B = \frac{0.005}{0.05}

  = 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

4 0
3 years ago
What kind of organic compound is propane
arsen [322]

Answer:

hydrocarbon

Explanation:

3 0
2 years ago
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