Flerovium at its ground state is solid. It has electron configuration of [Rn]5f¹⁴6d¹⁰7s²7p². The expected number of valence electrons in a flerovium atom is 2. A ground state is the most stable state of an atom at satndard temperature and pressure.
Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)
when we have STP conditions, we can use this conversion: 1 mol = 22.4 L
first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.
molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)
calculations:
Answer:
9.6 %
Explanation:
<u>Step 1: How to define percent error ? </u>
⇒ % error is the difference between a measured value and the known or accepted value
⇒Percent error is calculated using the following formula:
⇒%error = | Experimental value-theoretical value/theoretical value | x100%
⇔ this can be written as well as : error = (| Experimental value/ theoretical value | - | Theoretical value / Theoretical value | ) x100%
<u>Step 2: Calculate % error</u>
In this case, this means :
%error = ( |(4.45 cm - 4.06cm ) / 4.06cm | ) x100%
%error = 0.096 x100%
%error =9.6 %
Electronic configuration of the atom describes the arrangemnet of electrons in different shells and subshells ( sublevels).
Now , there are 4 types of sublevels: s, p , d and f . These sublevels have orbital which are spaces with high probability of having an electron and each orbital can have maximum 2 electrons.
Therefore,
s-sublevel has 1 orbital - it can have maximum 2 electrons.
p-sublevel has 3 orbitals - it can have maximum 6 electrons
d-sublevel has 5 orbitals - it can have maximum 10 electrons
f-sublevel has 7 orbitals - it can have maximum 14 electrons.
Hence, the acsending order of sublevels in terms of maximum number of electrons is:
<h2>s < p < d < f</h2>
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
