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ANEK [815]
2 years ago
12

Consider a gas in a container that can adjust its volume to maintain constant pressure. Suppose the gas is cooled. What happens

to the gas particles with the decrease in temperature? What happens to the volume of the container?
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
4 0

Answer:

The volume will also decrease.

Explanation:

This illustration clearly indicates Boyle's law.

Boyle's law states that the volume of a fixed mass of gas is directly proportional to the absolute temperature, provided the pressure remains constant. Mathematically, it is represented as:

V & T

V = KT

K = V/T

V1/T1 = V2/T2 =... = Vn/Tn

Where:

T1 and T2 are the initial and final temperature respectively, measured in Kelvin.

V1 and V2 are the initial and final volume of the gas respectively.

From the illustration above, the volume is directly proportional to the temperature. This implies that as the temperature increases, the volume will also increase and as the temperature decreases, the volume also will decrease.

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The formula of sodium oxide is Na2O
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Lavar el arroz y escurrirle el agua es un ejemplo de mezcla....
Tresset [83]

Answer:

me no oblo espenul

Explanation:

4 0
2 years ago
1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm
sashaice [31]
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm


Ideal gas equation: pV = nRT => n = pV / RT


R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm


=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol


2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type


X (+) + O2 (g) ---> X2O          or   


2 X(2+) + O2(g) ----> X2O2 = 2XO     or


4X(3+) + 3O2(g) ---> 2X2O3


 
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)



So, lets probe those 3 cases.


3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X


Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol


That does not correspond to any of the metal with valence 1+


So, now probe the case 2.



4) Case 2:


2moles X metal / 1 mol O2(g) = x / 0.01226 mol


=> x = 2 * 0.01226 = 0.02452 mol


And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol


That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.


4) Case 3


4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635 


atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol


That does not correspond to any metal.


Conclusion: the identity of the metallic element could be titanium.
5 0
3 years ago
Describe how you might determine the m/z and relative abundance of the ions contributing to the peak at 21.876 min
aliya0001 [1]

The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.

<h3>What is relative abundance?</h3>
  • The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
  • When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
  • Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.

To learn more about relative abundance with the given link

brainly.com/question/1594226

#SPJ4

8 0
1 year ago
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