All categories

2 years ago

15. An apparatus consists of a 4.0 dm3

Chemistry

2 answers:

Solnce55 [7]2 years ago

6 0

Answer:

i· Partial pressure of nitrogen gas is 219.429kPa and partial pressure of argon gas is 33.714kPa ·

ii· Total pressure of the gas mixture is 253.143kPa·

Explanation:

<h3>solution for i :</h3>

Assuming that the given gases to be ideal,

so by ideal gas equation

PV=nRT

where,

P is the pressure of the gas

V is the volume occupied by the gas

n is the number of moles of the gas

R is the ideal gas constant

T is the temperature of the gas

Actually partial pressure of each gas is the pressure of the gas exerted when it occupies complete volume

As the temperature of both gases are same, the mixing process is an isothermal process

PV=constant

Initially for nitrogen gas PV=803×4=3212

let P $x_{1}$ be the partial pressure of the nitrogen gas

(P $x_{1}$ )×14=3212

∴P $x_{1}$ =219.429kPa

∴Partial pressure of nitrogen gas is 219.429kPa

Initially for argon gas PV=47.2×10=472

let P $x_{2}$ be the partial pressure of the argon gas

(P $x_{2}$ )×14=472

∴P $x_{2}$ =33.714kPa

∴Partial pressure of argon gas is 33.714kPa

solution for ii :

Total pressure of the gas mixture will be the sum of the partial pressures of each gas as

Partial pressure of the gas=(total pressure of the mixture)×(mole fraction of the gas)

∴Total pressure=P $x_{1}$ +P $x_{2}$

=219.429+33.719

Total pressure =253.143kPa

igomit [66]2 years ago

6 0

Partial pressure : <u>P N₂ = 2.251 atm; P Ar = 0.3336 atm</u>

Total Pressure = <u>2.585 atm</u>

<h3>Further explanation </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated), among others

Boyle's law at constant T, P = 1 / V
Charles's law, at constant P, V = T
Avogadro's law, at constant P and T, V = n

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large{\boxed{\bold{PV=nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08205 l.atm / mol K

T = temperature, Kelvin

Find mol of each gas before valve is opened

1. mol N₂

P = 803 kPa = 7.925 atm

V = 4 L

T = 25 C = 298 K

$\rm n=\dfrac{7.925\times 4}{0.08205\times 298}\\\\n=1.289$

2. mol Ar

P = 47.2 kPa = 0.466 atm

V = 10 L

$\rm n=\dfrac{0.466\times 10}{0.08205\times 298}\\\\n=0.191$

Then :

V total = 14 L

Find partial pressure of each gas after valve is opened

1. P N₂

$\rm P_{N_2}=\dfrac{1.289\times 0.08205\times 298}{14}\\\\P_{N_2}=2.251\:atm$

2. P Ar

$\rm P_{Ar}=\dfrac{0.191\times 0.08205\times 298}{14}\\\\P_{Ar}=0.3336\:atm$

P total = P N₂ + P Ar

P total = 2.251 + 0.3336

P total = 2.585 atm

<h3>Learn more </h3>

Which equation agrees with the ideal gas law

brainly.com/question/3778152

brainly.com/question/1056445

Which law relates to the ideal gas law

brainly.com/question/6534096

You might be interested in

Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of

NeX [460]

Answer:

Molarity of HCl = 0.19 M

Explanation:

Moles of $Na_2CO_3$ :-

Mass = 0.246 g

Molar mass of $Na_2CO_3$ = 105.988 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{0.246\ g}{105.988\ g/mol}$

$Moles_{Na_2CO_3}= 0.002321\ mol$

According to the given reaction:-

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2O+CO_2$

1 mole of $Na_2CO_3$ react with 2 moles of HCl

0.002321 mole of $Na_2CO_3$ react with 2*0.002321 moles of HCl

Moles of HCl = 0.004642 moles

Given that volume = 24.3 mL

Also,

$1\ mL=10^{-3}\ L$

So, Volume = 24.3 / 1000 L = 0.0243 L

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.004642}{0.0243}$

<u>Molarity of HCl = 0.19 M </u>

<u></u>

7 0

2 years ago

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a

kvv77 [185]

The question is incomplete, the complete question is;

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75

Answer:

A nucleus with a A nucleus with a neutron:proton ratio of 1.49

A nucleus with a mass of 187 and an atomic number of 75

Explanation:

The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1

Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.

Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.

Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.

5 0

1 year ago

Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a

Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

Pressure = 1.50 *10^6 Pa

The pressure reduce to 0.950 *10^6 Pa

The temperature stays constant at 200 °C

Step 2: Calculate the volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.50 * 10^6 Pa

⇒with V1 = the initial volume = 25 L

⇒with P2 = the final pressure = 0.950 * 10^6 Pa

⇒with V2 = the final volume = TO BE DETERMINED

1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2

V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)

V2 = 39.5 L = 0.0395 m³

The final volume is 39.5 L = 0.0395 m³

3 0

2 years ago

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of

Alecsey [184]

Answer:

Without dark matter galaxies would loose an extreme amount of gas required to create stars.

Without dark matter the universe wont have as many galaxies clumped together forming larger versions of those galaxies. This would cause a change in the structure of the "skeleton" of the web.

(Hope this can help, I didn't do exactly as it is said to because that is your job)

Explanation:

Forbes gives somewhat of an explanation if you are curious.

(Ethan Siegal, "The Universe Would Be Very Different Without Dark Matter", Forbes)

3 0

1 year ago

Molecular bromine is 24 percent dissociated at 1600 k and 1.00 bar in the equilibrium br2 (

dlinn [17]

Br2 == 2Br

24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2

=> partial pressure of Br, P Br = 0.24 bar, and
partical pressure of Br2, P Br2 = 0.76 bar

kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758

3 0

2 years ago