Answer:
Part A: Hexacyanoferrate (III)
Part B: DiammintetraaquoCupperate (II)
Part C: Dichlorobis(ethylenediamine) Chromate (II)
Part D: Triaquocarbonylnickel (II) Sulphate
Part E: Potassium Dicarbonatedifluoroplatinate (II)
Explanation:
For naming the complex ions there is a specific rule
Nomenclature of the complex ions are as follow
- write a correct formulae
- Indicate the oxidation number of metal in the complex
- The oxidation number should write in the roman numeral in perenthasis after metal name
- Ligand named before the metal ion
- Ligan can be named in following order
* 1st negative, 2nd neutral, 3rd positive
* If there are more than 2 same charged ligand the write in
alphabetical order.
- Write prefix i.e di, tri, tetra for multiple monodentate ligands
- Anions name end at ido the replace the final name.
- Neutral ligands named as their usual name, but there are some exceptions such as
NH3 named as ammine
H2O names aqua or aquo
CO named ascarbonyl
NO named as nitrosyl
- If the complex is an anion, then name of the central atom will end in -ate, and its Latin name will be used except for mercury
- The name of full complex will end with cation or anion with separate word.
Keeping the rules in mind the complexes named as following.
_________________________
Part A:
[Fe(CN)₆]³⁻
Name of the Complex : Hexacyanoferrate (III)
___________________
Part B:
[Cu(NH₃)₂(H₂O)₄]²⁺
Name of the Complex : DiammintetraaquoCupperate (II)
_______________________
Part C
CrCl₂(en)₂
Name of the Complex : Dichlorobis(ethylenediamine) Chromate (II)
________________________
Part C
[Ni(H₂O)₃(CO)]SO₄
Name of the Complex : Tetraaquocarbonylnickel (II) Sulphate
______________________
Part E
K₄[Pt(CO₃)₂F₂]
Name of the Complex : Potassium Dicarbonatedifluoroplatinate (II)
Let's assume that both He and N₂ have ideal gas behavior.<span>
Then we can use ideal gas law,
PV = nRT
Where, P is the pressure of gas, V is the volume,
n is moles of gas, R is universal gas constant and T is the temperature in
Kelvin.
<span>The </span>P <span>and </span>V <span>are </span>same<span> for the
both gases.</span>
R is a
constant.
The only variables are n and T.
<span>Let's say temperature of </span>He<span> <span>is </span></span>T</span>₁<span> <span>and temperature of </span></span>N₂<span> <span>is </span></span>T₂.<span>
n = m/M<span> where n is
moles, m is mass and M is molar mass.</span>
Molar mass of He is 4 g/mol and molar mass of N₂ is 28 g/mol</span><span>
<span>Since mass (m) of both gases are same,</span>
moles of He = m/4
moles of N₂ = m/28</span><span>
Let's apply the ideal gas equation for both gases.
For He gas,
PV = (m/4)RT₁ </span>(1)<span>
For N</span>₂ gas,<span>
PV = (m/28)RT₂<span> </span></span> (2)<span>
(1) = (2)
</span><span>(m/4)RT₁ =
(m/28)RT₂</span> <span>
T₁/4
= T₂/28</span><span>
T₁ = T₂/7</span><span>
<span> </span>7T</span>₁ = T₂<span>
Hence, the
temperature of N</span>₂<span> gas is higher by 7
times than the temperature of He gas.</span>
Answer:
0.1 is the retention factor.
Explanation:
Distance covered by solvent ,
Distance covered by solute or ion,
Retention factor
is defined as ratio of distance traveled by solute to the distance traveled by solvent.


0.1 is the retention factor.
Answer:
Here's what I get
Explanation:
(a) Mass % of each ion

I used this template to calculate the percentages in the table below.
<u> Ion Abundance Mass % </u>
Na⁺ 10 560 30.72
K⁺ <u> 380 </u> <u> 1.11 </u>
Alkali metals 10 940 31.82
Mg²⁺ 1 270 3.69
Ca²⁺ <u>400 </u> <u> 1.16 </u>
Alkaline earth metals <u>1 670 </u> <u> 4.86 </u>
Total metal ions 12 610 36.68
Cl⁻ 18 980 55.20
SO₄²⁻ 2 650 7.71
HCO₃⁻ 140 0.41
Anions <u>21 770 </u> <u> 63.32
</u>
TOTAL 34 380 100.00
(b) 30.72 % of the total mass is sodium ion.
(c) Alkaline earth metals vs alkali metals

The mass percent of alkali metal ions is 6.55 times that of alkaline earth metal ions.
(d) The mass of anions is greater than that of cations.