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Digiron [165]
2 years ago
4

14.0 moles of gas are in a 3.00 LL tank at 23.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas und

er these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2a=2.300L2⋅atm/mol2 and b=0.0430 L/molb=0.0430 L/mol. Express your answer with the appropriate units.
Chemistry
1 answer:
Yakvenalex [24]2 years ago
7 0

Answer:

The difference in pressure = 21.59 atm

Explanation:

Step 1: Data given

Moles of gas = 14.0 moles

Volume = 3.00 L

Temperature = 23.3 °C

The van der Waals constants for methane are a=2.300L²*atm/mol² and b=0.0430 L/mol

Step 2: Calculate pressure

p*V = n*R*T

⇒ p = the pressure = TO BE DETERMINED

⇒ V = the volume of the tank = 3.00 L

⇒ R = the gas constant = 0.08206 L*atm/mol*K

⇒ T = the temperature = 23.3 °C = 296.45 K

⇒ n = the number of moles gas = 14.0 moles

p = (n*R*T)/V

p = (14.0 *0.08206 * 296.45)/3.00

p = 113.52 atm

Step 3: Calculate the pressure via the van der waals equation

The Van der waals equation is (p + n²a/V²)*(V-nb) = nRT 

⇒p = To be determined

⇒ n =the number of moles = 14.0 moles

⇒a=2.300 L²*atm/mol²

⇒b=0.0430 L/mol

(p + 14.0²*2.300 / 3.00²) * (3.00 - 14.0*0.0430) = 14.0*0.08206*296.45

(p + 14.0²*2.300 / 3.00²) * (3.00 - 14.0*0.0430) = 340.57

(p + 14.0²*2.300 / 3.00²) * 2.398 = 340.57

(p + 14.0²*2.300 / 3.00²) = 142.02

(p + 50.09) = 142.02

p = 91.93 atm

The difference in pressure = 113.52 atm - 91.93 atm = 21.59 atm

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bazaltina [42]

Answer:

Option 4th, TRUE . Its pH becomes equal to 7.

Explanation:

[H₃O⁺] = [OH⁻]

pH = pOH → 7

[H₃O⁺] > [OH⁻]  → pH < 7 ⇒ acidic solution

[H₃O⁺] < [OH⁻]  → pH > 7 ⇒ basic solution

pH = 0 does not exist.

7 0
2 years ago
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A gram of gasoline produces 45.0kj of energy when burned. gasoline has a density of 0.77/gml . how would you calculate the amoun
babymother [125]
Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:

48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ

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5 0
2 years ago
Kaia, a chemical engineering graduate, has documented all titration procedures in her project report. She refers to this report
vagabundo [1.1K]

Answer:

The correct option is;

d. Explicit knowledge

Explanation:

Explicit knowledge is the knowledge that can be easily articulated documented stored in a retrieval system accesses, transmitted and shared with others

Tacit knowledge is the skill developed by an individual based on actual experience such that such knowledge comprise of both facts and perspectives

Hence explicit knowledge and tacit knowledge are complementary

The operations performed by Kaia include documentation, storing in a retrieval system (her project report) and accessing what she documented, this is an example of explicit knowledge.

7 0
2 years 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
A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O
Alexxandr [17]
C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O  produces 24 + 4 + 32 g C2H4O2    (theoretical)
 46 g produces 60g 

60 g C2H4O2 is produced from 46g C2H6O
1g      .     .................................46/60 g 
 700g     .................................    (46/60) * 700  Theoretically

But as the yield is only 7.5% 

the required amount is    ((46/60) * 700 ) / 0.075 =  7155.56 g 

=  7.156 kg to nearest gram.  Answer




8 0
2 years ago
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