Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Answer:
Explanation:
First digit of the 2p^3 gives you value of n, in this case its = 2, So, n= 2
Second alphabet gives you the value of l,
l=0 =s
l=1 =p
l=3=d
l=4=f
since "p" is the alphabet in 2p^3, so in your case lt shoudlbe = 1 right?
ml= -l to +l , that is -1, 0, +1
Ms= +1/2 or -1/2 alaways remains same foe evrything.
Anaphase 1 is when centromeres divide
