I’m confused what are you asking exactly?
O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol
Answer:
The concentration of the solution will be much lower than 6M
Explanation:
To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.
From
n= CV
n = number of moles m/M( m= mass of solid, M= molar mass of compound)
C= concentration of substance
V= volume of solution
m=120g
M= 40gmol-1
V=500ml
120/40= C×500/1000
C= 120/40× 1000/500
C=6M
This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.
This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.
All the following are equal to Avogadro's number EXCEPT a. the number of atoms of bromine in 1 mol Br₂.
1 mol Br₂ contains Avogadro’s number of molecules of Br₂.
However, each molecule contains two atoms of Br, so there are
<em>2 × Avogadro’s number of Br atoms </em>in 1 mol Br₂.
Answer:
The answer is D.) Beaker. Hope this helps! :)
