Experiments test the scientists' ideas.
Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.
For the compound, 
the electronic configuration of the atoms, carbon and hydrogen are:
Carbon (atomic number=6): In ground state= 
In excited state: 
Hydrogen (atomic number=1): 
All the bonds in the compound is single bond(
-bond) that is they are formed by head on collision of the orbitals.
The structure of the compound is shown in the image.
The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.
In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in of hydrogen will overlap to the 2p^{3}-orbital of carbon.
Thus, the hybridization of Hydrogen is 
-hybridization and the hybridization of Carbon is 
-hybridization.
The hybridization of each atom is shown in the image.
Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.
Rb: [Kr] 5s¹ - 1 valence electron
Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons
Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons
I: [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons
In: [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons
Rank from most to fewest valence electrons:
Xe > I > Sb > In > Rb
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
