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Firlakuza [10]
3 years ago
10

Help please. 15.0 moles of gas are in a 3.00L tank at 23.4∘C . Calculate the difference in pressure between methane and an ideal

gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol..
Chemistry
1 answer:
CaHeK987 [17]3 years ago
3 0
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT.  Substituting  a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
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Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

8 0
3 years ago
What is the identity of the element which had the following electron configuration? 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​
kati45 [8]

Answer:

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

Explanation:

Step 1: Data given

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​

has 25 electrons.

This element has 2 electrons on the first shell, 8 electrons on the second shell, 13 electrons on the third shell and 2 electrons on the outer shell (valence electrons).

This means this element is part of group VII.

The element with 25 electrons, we can find on the periodic table, with atomic number 25.

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 ​ is manganese (25Mn).

5 0
3 years ago
If an element only partially conducts electricity, it would be part of the _______ category. *
dmitriy555 [2]
It would be in the transition metals
5 0
3 years ago
When many atoms are split in a chain reaction, a large explosion occurs. This is an example of what type of energy conversion? (
lara31 [8.8K]

Answer:

I think is chemical and nuclear

8 0
3 years ago
Read 2 more answers
What is the speed of a man walking for 30 meters and he walked for 5 seconds ​
aliya0001 [1]

Answer:

<h2>6 m/s</h2>

Explanation:

The speed of the man can be found by using the formula

v =  \frac{d}{t}  \\

d is the distance

t is the time taken

From the question we have

v =  \frac{30}{5}  \\

We have the final answer as

<h3>6 m/s</h3>

Hope this helps you

8 0
3 years ago
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