Answer:
28 g/mol, N2
Explanation:
Given data:
Volume of gas = 5.0 L
Mass of gas = 6.3 g
Pressure = 1 atm
Temperature = 273 K
Molar mass of gas = ?
Solution:
We will calculate the density first.
d = mass/ volume
d = 6.3 g/ 5.0 L
d = 1.26 g/L
Molar mass:
d = PM/RT
M = dRT/P
M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm
M = 28 g/mol
Molar mass of N₂ is 28 g/mol thus given gas is N₂.
Answer:
Francium (Fr)
Explanation:
Looking at the periodic table, Francium has 7 energy levels as it is in the 7th Period and is in the 1st Group (meaning it has one electron in the outermost shell) which suggests that it has one valence electron.
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Answer: The concentrations of 
at equilibrium is 0.023 M
Explanation:
Moles of = 
Volume of solution = 1 L
Initial concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of at equilibrium is 0.023 M
Answer: assume pathogens are present and treat the samples accordingly
Explanation:
When investigators are unable to conclusively ascertain the source of a biological sample found at a crime scene, the correct thing to do is to treat it as if pathogens are present in it and handle it according to set rules on how to handle pathogens.
This is done to ensure that if a pathogen is indeed present, it would not cause a health emergency by infecting those who come in contact with the samples at the scene.
The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is
2 g/mol + 3 g/mol = 5 g/mol AB
for the case of A2B3
A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol
therefore A2B3 = 13 g/mol
