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3 years ago

Are controlled substances and illegal drugs the same thing?

Chemistry

2 answers:

Reil [10]3 years ago

8 0

Well i do think they're the same.

SVETLANKA909090 [29]3 years ago

4 0

No: A controlled substance is a chemical compound that is "controlled," per say, by the government. Access to these substances is limited, hence the use of prescriptions for potentially addictive pharmaceuticals.
An illegal drug is a substance deemed harmful, both to the body of one using said substance and those in the society around it, and its use is, therefore, not allowed under the law.

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Which element has this electron configuration? <br> astatine<br> bromine<br> chlorine<br> fluorine

oksano4ka [1.4K]

Unfortunately you did not specify the electronic configuration in the question, however since one of the answers must be a halogen, i took the liberty to attach an image with the configuration (both the simple numeric and spdf form) for all the halogen and all you have to do is match the electronic configuration you have in your question to the one in the table attached and you can then deduce the answer.

Hope this helps.

7 0

3 years ago

Read 2 more answers

What is the number of the lowest energy level that contains an f sublevel?

forsale [732]

Answer: n = 4.

Explanation:

1) The main energy level is represented by the principal (main) quantum number, n. It can be 1, 2, 3, 4, 5,6, or 7. It is the period number in the periodic table.

2) The next quantum number is the Azimuthal quantum number (ℓ), also known as angular quantum number. It indicates the subshell (sublevel) or type of orbital.

This quantun number may be:

s, p, d, or f.

3) For n = 1, there is only one ℓ number: 0; which is the orbital s

For n = 2, there are two possible ℓ numbers: 0 and 1; which are the orbitals s and p.

For n = 3, there are three possible ℓ numbers: 0, 1, and 2, which are the orbitals s, p, and d:

For n = 4, there are four possible ℓ numbers:0, 1, 2, and 3; which are the orbitals s, p, d, and f.

So, the f sublevel appears first time when n = 4, i.e the number of the lowest energy level that contains an f sublevel is 4.

5 0

3 years ago

If 125.0g of nitrogen is reacted with 125.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

MakcuM [25]

Answer:

Hydrogen is the excess reactant

Nitrogen is the limiting reactant

151.6g is theoretical yield

Explanation:

The reaction of N₂ with H₂ to produce NH₃ is:

N₂ + 3H₂ → 2NH₃

To find theoretical yield we need to determine limiting reactant with the moles of each gas as follows:

Nitrogen -Molar mass: 28g/mol-

125.0g * (1mol / 28g) = 4.46 moles

Hydrogen -Molar mass: 2g/mol-

125.0g * (1mol / 2g) = 62.5 moles of hydrogen

For a complete reaction of 4.46 moles of N2 there are needed:

4.46 moles N2 * (3moles H2 / 1mol N2) = 13.38 moles of hydrogen

As there are 62.5 moles of hydrogen:

<h3>Hydrogen is the excess reactant</h3><h3>Nitrogen is the limiting reactant</h3><h3 />

With nitrogen, the limiting reactant, we determine theoretical moles (Assuming 100% of the reaction occurs) and theoretical yield (In mass):

4.46 moles N2 * (2moles NH3 / 1mol N2) = 8.92 moles of ammonia

As molar mass of ammonia is 17g/mol:

8.92 moles of ammonia * (17g/mol) =

<h3>151.6g is theoretical yield</h3>

5 0

3 years ago

What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×

Ira Lisetskai [31]

$^{10}_5 B$
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

$5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\
8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\
16.7375 \times 10^{-27}= \\
1.67375 \times 10^{-26} \approx \\
1.67 \times 10^{-26}$

The answer is A. 1.67 × 10⁻²⁶ kg.

8 0

3 years ago

Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th

Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ = C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

6 0

3 years ago