HEY THERE!!
For a given principal quantum number or n, the corresponding angular quantum number or l is equivalent to a range between 0 and (n-1) .
This means that the angular quantum number for a principal quantum number of 2 is equivalent to:
l = 0 -> (n-1) = 0 -> (2-1) = 0 -> 1
So the answer is 0, 1
HOPE IT HELPED YOU.
Answer:
0.21 M. (2 sig. fig.)
Explanation:
The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.
How many moles of NaOH in the original solution?
,
where
is the number of moles of the solute in the solution.
is the concentration of the solution. 
for the initial solution.
is the volume of the solution. For the initial solution, 
for the initial solution.
.
What's the concentration of the diluted solution?
.
- is the number of solute in the solution. Diluting the solution does not influence the value of .
for the diluted solution. - Volume of the diluted solution:
.
Concentration of the diluted solution:
.
The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
Answer:
0.208mole of CO2
Explanation:
First, let us calculate the number of mole of HC3H3O2 present.
Molarity of HC3H3O2 = 0.833 mol/L
Volume = 25 mL = 25/100 = 0.25L
Mole =?
Mole = Molarity x Volume
Mole = 0.833 x 0.25
Mole of HC3H3O2 = 0.208mole
Now, we can easily find the number of mole of CO2 produce by doing the following:
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
From the equation,
1mole of HC2H3O2 produced 1 mole of CO2.
Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2
From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O=
= 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces 
moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.
