Electrons in an atom can be classified as core electrons and valence electrons. Valence electrons are those electrons which are present in valence shell and participates in bond formation. While, Core electrons are all remaining electrons which are not present in valence shell, hence not take part in bonding.
Atomic number of Selenium (Se) is 34 hence it has 34 electrons with following electronic configuration;
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
From electronic configuration it is found that the valence shell is 4, and the number of electrons present in valence shell are 6. So,
Core Electrons = Total Electrons - Valence Electrons
Core Electrons = 34 - 6
Core Electrons = 28
Result:
There are 28 core electrons in Selenium.
<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 4.53 V.
Answer:
1.0 M
Explanation:
Reaction equation;
KOH(aq) + HCl(aq) -----> KCl(aq) + H2O(l)
Concentration of acid CA = ?
Concentration of base CB = 1.0 M
Volume of base VB = 25.60 - 0.50 = 25.1 ml
Volume of acid VB = 25.0 ml
Number of moles of acid NA = 1
Number of moles of base NB =2
CAVA/CBVB =NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 1 * 25.1 * 1/25.0 *1
CA = 1.0 M
The formula for pH given the pKa and the concentrations
are:
pH = pKa + log [a–]/[ha]
<span>
Therefore calculating:</span>
3.75 = 3.75 + log [a–]/[ha]
log [a–]/[ha] = 0
[a–]/[ha] = 10^0
<span>[a–]/[ha] = 1</span>
Answer:
I think cold front if not than its c
