Answer:
Thomson placed two magnets on either side of the tube, and observed that this magnetic field also deflected the cathode ray. The results of these experiments helped Thomson determine the mass-to-charge ratio of the cathode ray particles, which led to a fascinating discovery, minus the mass of each particle was much, much smaller than that of any known atom. Thomson repeated his experiments using different metals as electrode materials, and found that the properties of the cathode ray remained constant no matter what cathode material they originated from. From this evidence, Thomson made the following conclusions:
The cathode ray is composed of negatively-charged particles.
The particles must exist as part of the atom, since the mass of each particle is only ~1/2000 the mass of a hydrogen atom.
These subatomic particles can be found within atoms of all elements.
While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of the electron disproved the part of Dalton's atomic theory that assumed atoms were indivisible. In order to account for the existence of the electrons, an entirely new atomic model was needed.
Explanation:
a) before addition of any KOH :
when we use the Ka equation & Ka = 4 x 10^-8 :
Ka = [H+]^2 / [ HCIO]
by substitution:
4 x 10^-8 = [H+]^2 / 0.21
[H+]^2 = (4 x 10^-8) * 0.21
= 8.4 x 10^-9
[H+] = √(8.4 x 10^-9)
= 9.2 x 10^-5 M
when PH = -㏒[H+]
PH = -㏒(9.2 x 10^-5)
= 4
b)After addition of 25 mL of KOH: this produces a buffer solution
So, we will use Henderson-Hasselbalch equation to get PH:
PH = Pka +㏒[Salt]/[acid]
first, we have to get moles of HCIO= molarity * volume
=0.21M * 0.05L
= 0.0105 moles
then, moles of KOH = molarity * volume
= 0.21 * 0.025
=0.00525 moles
∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525
and when the total volume is = 0.05 L + 0.025 L = 0.075 L
So the molarity of HCIO = moles HCIO remaining / total volume
= 0.00525 / 0.075
=0.07 M
and molarity of KCIO = moles KCIO / total volume
= 0.00525 / 0.075
= 0.07 M
and when Ka = 4 x 10^-8
∴Pka =-㏒Ka
= -㏒(4 x 10^-8)
= 7.4
by substitution in H-H equation:
PH = 7.4 + ㏒(0.07/0.07)
∴PH = 7.4
c) after addition of 35 mL of KOH:
we will use the H-H equation again as we have a buffer solution:
PH = Pka + ㏒[salt/acid]
first, we have to get moles HCIO = molarity * volume
= 0.21 M * 0.05L
= 0.0105 moles
then moles KOH = molarity * volume
= 0.22 M* 0.035 L
=0.0077 moles
∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5
when the total volume = 0.05L + 0.035L = 0.085 L
∴ the molarity of HCIO = moles HCIO remaining / total volume
= 8 x 10^-5 / 0.085
= 9.4 x 10^-4 M
and the molarity of KCIO = moles KCIO / total volume
= 0.0077M / 0.085L
= 0.09 M
by substitution:
PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)
∴PH = 8.38
D)After addition of 50 mL:
from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.
the molarity of KCIO = moles KCIO / total volume
= 0.0105mol / 0.1 L
= 0.105 M
when Ka = KW / Kb
∴Kb = 1 x 10^-14 / 4 x 10^-8
= 2.5 x 10^-7
by using Kb expression:
Kb = [CIO-] [OH-] / [KCIO]
when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]
Kb = [OH-]^2 / [KCIO]
2.5 x 10^-7 = [OH-]^2 /0.105
∴[OH-] = 0.00016 M
POH = -㏒[OH-]
∴POH = -㏒0.00016
= 3.8
∴PH = 14- POH
=14 - 3.8
PH = 10.2
e) after addition 60 mL of KOH:
when KOH neutralized all the HCIO so, to get the molarity of KOH solution
M1*V1= M2*V2
when M1 is the molarity of KOH solution
V1 is the total volume = 0.05 + 0.06 = 0.11 L
M2 = 0.21 M
V2 is the excess volume added of KOH = 0.01L
so by substitution:
M1 * 0.11L = 0.21*0.01L
∴M1 =0.02 M
∴[KOH] = [OH-] = 0.02 M
∴POH = -㏒[OH-]
= -㏒0.02
= 1.7
∴PH = 14- POH
= 14- 1.7
= 12.3
Answer:
atoms of isotopes of an element have different numbers of neutrons.
Explanation:
isotopes have the same number of protons, but different numbers of neutrons.
Answer:
The equivalent weight of M is approximately 31.8 g
The equivalent weight of N is approximately 27.98 g
Explanation:
The given parameters are;
The percentage of the the metal M in in the chloride = 47.25%
Where by the chemical formula for the metal chloride is MClₓ, we have;
47.25% of the mass of MClₓ = Mass of M = W
Therefore, we have;
0.4725 × (W + 35.5·x) = W
0.4725·W + 0.4725×35.5×x = W
W - 0.4725·W = 16.77·x
0.5275·W = 16.77·x
W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M
The equivalent weight of M = 31.799 ≈ 31.8 g
Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g
The equivalent weight of N = 27.98 g.
1. True 2. False 3.false 4. False 5. True
