It’s C. 0.31 atm
I hope this helped out! Have a nice day :)
First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5
The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
Answer:
0.0738 M
Explanation:
HNO3 +LiOH = LiNO3 + H2O
Number of moles HNO3 = number of moles LiOH
M(HNO3)*V(HNO3) = M(LiOH)*M(LiOH)
M(HNO3)*50.0mL = 0.100M*36.90 mL
M(HNO3) = 0.100*36.90/50.0 M = 0.0738 M
From the equation;
4 Al + 3 O2 = 2 Al2O3
The mole ratio of Oxygen is to Aluminium hydroxide is 3:2.
Therefore; moles of Al2O3 is
(0.5/3 )× 2 = 0.333 moles
Therefore; The moles of aluminium oxide will be 0.333 moles
Answer:
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