Answer: n= 3, l= 0 , m= 0 and s=[/tex]\frac{-1}{2}[/tex]
Explanation:
Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic quantum number : It describes the orientation of the orbital. It is represented as 'm'. The value of m ranges from -l to +l.
Spin quantum number: It describes the spin of the electron. It is represented as 's'. The value of s can be +1/2 or -1/2
Thus electron enters 3s orbital , thus n= 3, l= 0 , m= 0 and s=[/tex]\frac{-1}{2}[/tex]
Potential energy can be calculated by the formula Pe=mgh. Plug in your values:
Pe=mgh
Pe=(6 kg)(9.8m/s^2)(100 m)
Pe=5880 kg x m^2/s^2, or 5880 Joules
Answer:
D. The chemical formula
Explanation:
For example, in the compound KCl, we know that there are two elements present because you can see it in the chemical formula. We know that KCl consists of potassium and chloride ions.
Answer : The most likely happens during this reaction is, Oxidation-reduction
Explanation :
The balanced reaction will be,
In this reaction, neutral iron loses 3 electrons and oxidizes in (+3) state, 
and neutral oxygen gains 2 electrons and reduces in (-2) state,
When iron react with oxygen gas to give iron oxide. This process is known as iron rusting. During the reaction, oxidation-reduction process occurs.
Oxidation : It is a type of chemical reaction in which a substance loses its electrons. Or we can say that in oxidation, the oxidation number increases.
Reduction : It is a type of chemical reaction in which a substance gains its electrons. Or we can say that in reduction, the oxidation number decreases.
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
