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1 year ago

Because amines are basic, they can often accept a proton. draw the protonated structure of n-propylamine

1 answer:

3 0

Answer is: CH₃-CH₂-CH₂-NH₃⁺.
Amine group (-NH₂) has neutral charge, bet when accepts one proton (H⁺) has positive charge.
Amines are compounds that contain a basic nitrogen atom with a lone pair of electrons. Amines are derivatives of ammonia, where is one or more hydrogen atoms replaced by a substituent such as an alkyl or aryl group (in this compound propyl group).

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An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial

andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state ) P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm

P₂= 2.25 atm

8 0

1 year ago

A pharmacist–herbalist mixed 100 g lots o St. John’s wort containing the ollowing percentages o the active component hypericin:

Anna [14]

Answer:

strength of hypericin in mixture = 0.42 %

Explanation:

given data

each lot = 100 g

active component hypericin = 0.3%, 0.7%, and 0.25%

solution

we get here percent strength o hypericin in the mixture that is

Hypericin contribution lot 1 = $\frac{0.3}{100}$ × 100

Hypericin contribution lot 1 = 0.3 g

and

Hypericin contribution lot 2 = $\frac{0.3}{100}$ × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = $\frac{0.25}{100}$ × 100

Hypericin contribution lot 3 = 0.25 g

total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g

total 300 g mixture of hypericin = 1.25 g

so here percent strength o hypericin in mixture is

strength of hypericin in mixture = $\frac{1.25}{300}$ × 100

strength of hypericin in mixture = 0.42 %

5 0

1 year ago

Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 × 10−13 at some temperature, compute

Artyom0805 [142]

Since Kw= [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²
So,
3.31x10⁻¹³ = [H⁺]²
Take the square root= 5.75x10⁻⁷
Then take the negative log to find the pH:
-log(5.75x10⁻⁷) = 6.25

5 0

1 year ago

A 1.50-liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25°C. Without changing th

Aneli [31]

Hope this helps you.

3 0

2 years ago

Read 2 more answers

How many simple distillation columns are required to purify a stream containing five components into ﬁve 'pure"products? Sketch

Nitella [24]

Answer: one simple distillation column is required to separate the stream into five pure products. With four different flat bottom flask, for collection of the distilled products

Explanation: simple distillation works with the difference in boiling points of the liquid to be separated. For the separation of five different constituent to be possible, we have to know the boiling points of the constituents.

For your understanding, let's define constituents in the liquid to be A, B, C, D, E. And the boiling points increases respectively. Start by heating the liquid to the boiling point of A to extract A. After a while check if the constituents A is still dropping in the flat bottom flask, if it has stopped dropping, it simply means that we have extracted all A constituents in the liquid, label the Flask A. Get another flask to extract constituent B.

Heat the mixture to the boiling point of B, after a while check if constituent B is still dropping in the flat bottom flask, if it has stopped dropping,it means that we have extracted all B constituent in the liquid, label the Flask B. Get another flask for C.

Repeat the same process for C and D.

After Extracting D we don't need to distillate E because we already have a pure form of E inside to the conical flask.

SEE PICTURE TO UNDERSTAND WHAT A SIMPLE DISTILLATION LOOKS LIKE

3 0

1 year ago