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:
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Explanation:
100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)
Hope this helps~
Answer:
Option A. KCl (aq)
Option D. Mg(OH)₂(s
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
MgCl₂(aq) + KOH(aq) —>
In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
KOH(aq) —> K⁺(aq) + OH¯(aq)
MgCl₂(aq) + KOH(aq) —>
Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)
MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)
Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)
Thus, option A and D gives the correct answer to the question.
Answer:
H₂: 0.48, N₂: 0.43; Ar: 0.09
Explanation:
First of all, sum all the pressures to know the total pressure in the mixture.
434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr
Mole fraction = Pressure gas / Total Pressure
Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48
Mole Fraction N₂: 389.9 /901.8 Torr =0.43
Mole Fraction Ar: 77.9 /901.8 Torr = 0.09
Remember: <u>SUM OF MOLE FRACTION = 1</u>