What is the most reactable element in periodic table
2 answers:
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Answer:
13.
a.Atomic number of cadmium is 48
b.Number of protons = 48
c.Mass number = 115 u
d.Number of neutrons=67
e.Isotopic Symbol=
14. Elements which are isotopes are:
- A ,C and E
- B and D
Explanation:
13.
a.Atomic number(Z):The number of protons present in the nucleus of the atom. Every element have fixed characteristic atomic number which decides its chemical property.It is represented by Z
for example : hydrogen atomic no. is 1 ,He is 2
So Cadmium has attained 48 atomic number.
b.Number of protons = Atomic number of the element
The atomic number of Cadmium is 48. Hence
Number of protons in Cadmium=48
c.Mass number(A) = Total number of protons and neutrons present in the nucleus of the atom.It is represented by A
Mass number = number of proton + number of neutron.....(1)
Cadmium has mass number of 115
d.Using equation (1),
Number of neutron = mass number - number of proton
Number of protons in Cadmium=48
Cadmium has mass number = 115
Number of neutron = 115 - 48
=67
e.The format of symbolic representation of elements is given below
where X=symbol of element
Z= Atomic number
A= mass number
For Cadmium , X = Cd,Z = 48 , A =115
Symbol=
(Since this is isotope of cadmium hence it is isotopic symbol)
14.
Isotopes: These are the species which have same number of protons(or atomic number) but different number of neutrons(mass number).
- A ,C and E have same number of proton = 22 but different mass number. mass number :A =55,C=56 and E=58
So A ,C and E are isotopes
- B and D have same number of proton = 25 but different mass number
mass number: B=55 ,D=57
B and D are isotopes
3.08 is the pOH of a 0.175 M aqueous solution of .
0.215% is the per cent ionization of a 0.325 M aqueous solution of
pH is a measure of how acidic/basic water is.
A)
→
Kb = 4.5 x
Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).
concentration of (NH₄⁺) = concentration of (OH⁻) = x.
x² = Kb x concentration of (NH₃)
x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.
x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)
= 8.367 × 10⁻⁴
pOH = -log(c(OH⁻))
=- log ( 8.367 × 10⁻⁴)
= 3.08
B)
Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.
Concentration of (NX₃) = 0.325 M.
Kb = 4.5 x 10⁻⁶.
[NX₃H⁺] = [OH⁻] = x.
[NX₃] = 0.325 M - x.
Kb = [NX₃H⁺] x [OH⁻] ÷ [NX₃].
4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).
x = 0.0007 M.
Per cent of ionization:
α = 0. 0007 M ÷ 0. 325 M x 100%
= 0.215%.
Hence,
3.08 is the pOH of a 0.175 M aqueous solution of .
0.215% is the per cent ionization of a 0.325 M aqueous solution of
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