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sergij07 [2.7K]
2 years ago
15

A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres

sure is 1.350 atm . Part A What is the partial pressure of argon, PAr, in the flask?
Chemistry
1 answer:
tatiyna2 years ago
4 0

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 m^{3}

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 \frac{J}{kg k}

From ideal gas equation P V = m RT

⇒ P = \frac{m R T}{V}

Put all the values in above formula we get

⇒ P = \frac{0.00115}{0.001} × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa

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Explain and illustrate the notation for distinguishing between the different p orbitals in a sublevel.
masha68 [24]

Here, the three different notation of the p-orbital in different sub-level have to generate

The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number m_{l} depends upon the value of l, which are -l to +l.

Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as p_{x}, p_{y} and p_{z} in space.

The three p-orbital can be distinguish by the quantum numbers as-

For 2p orbitals (principal quantum number is 2)

1) n = 2, l = 1, m = -1

2) n = 2, l = 1, m = 0

3) n = 2, l = 1, m = +1

Thus the notation of different p-orbitals in the sub level are determined.  

6 0
1 year ago
Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
stira [4]

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

8 0
2 years ago
Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
Basile [38]

Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{128\ g}{32.04\ g/mol}

Moles\ of\ methanol = 3.995\ mol

Given, Mass of water = 108 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{108\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.995\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}

Mole\ fraction\ of\ methanol=\frac{3.995}{3.995+5.995}=0.39989

<u>Mole fraction of methanol will be closest to 4.</u>

5 0
2 years ago
Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2
Brilliant_brown [7]

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and  equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>

<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>

<em />

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}

And Q, is:

Q = \frac{[NH_3]^2}{[N_2][H_2]^3}

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}

<h3>Q = 1.6x10⁻⁷</h3>

As Q < K,

<h3>The chemical system will shift to the right in order to produce more NH₃</h3>

Learn more about chemical equililbrium in:

brainly.com/question/24301138

8 0
1 year ago
The U.S. Mint produces a dollar coin called the American Silver Eagle that is made of nearly pure silver. This coin has a diamet
Rudik [331]

Answer:

The value of the silver in the coin is 35.3 $

Explanation:

First of all, let's calculate the volume of the coin.

2π . r² . thickness = volume

r = diameter/2

r = 41 mm/2 = 20.5 mm

2 . π . (20.5 mm)² .  2.5 mm = 6601 mm³

Now, this is the volume of the coin, so we must find out how many grams are on it.

6601 mm³ / 1000 = 6.60 cm³

Let's apply density.

D = Mass / volume

10.5 g/cm³ = mass /6.60 cm³

10.5 g/cm³ . 6.60 cm³ = mass

69.3 g = mass

Each gram has a cost of 0.51$

69.3 g . 0.51$ = 35.3 $

7 0
1 year ago
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