Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.
Answer : The molarity of calcium ion on the original solution is, 0.131 M
Explanation :
The balanced chemical reaction is:
When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.
First we have to calculate the moles of 
Given:
Mass of = 0.524 g
Molar mass of = 100 g/mol
Now we have to calculate the concentration of
Now we have to calculate the concentration of calcium ion.
As, calcium carbonate dissociate to give calcium ion and carbonate ion.
So,
Concentration of calcium ion = Concentration of = 0.131 M
Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
<span> Au</span>₂(SeO₄)₃
O = -2 × 4 = -8
Se = + 6
So,
(+6 - 8) = -2
Means (SeO₄) contains -2 charge, Now multiply -2 by 3
-2 ₓ 3 = -6
Means,
Au₂ + (-6) = 0
Au₂ = +6
Or,
Au = 6 / 2
Au = +3
Result:
Au = +3
Se = +6
O = -2
Ni(CN)₂
Cyanide (CN⁻) contains -1 charge,
So,
N = -3
C = +2
Then,
Ni + (-1)₂ = 0
Ni - 2 = 0
Or,
Ni = +2
Result:
N = -3
C = +2
Ni = +2
Solar Cycle 25 is now underway and expected to peak with 115 sunspots in July 2025.
