Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
Answer:
1,070.41 grams of DDT will be formed .Explanation:
1) 
Moles of chlorobenzene = 
According to reaction, 2 moles of chloro benzene reacts with 1 mole of chloral . Then 10.64 moles of chloro benzene will react with :
of chloral
2) Moles of chloral = 
According to reaction, 1 moles of chloral reacts with 2 mole of chlorobenzene . Then 3.0915 moles of chloral will react with :
of chloro benzene
As we can see that chloral is in limiting amount and chloro benzene is in excessive amount. So, amount of DDT will depend upon amount of chloral.
According to reaction, 1 mole chloral gives 1 mole DDT.Then 3.0195 moles of chloral will give :

Mass of 3.0195 moles of DTT :
3.0195 mol × 354.5 g/mol = 1,070.41 g
1,070.41 grams of DDT will be formed .
Decomposition, because the oil eventually breaks down bc of bacterias and evaporate into the ocean!
Answer:
t = 7.58 * 10¹⁹ seconds
Explanation:
First order rate constant is given as,
k = (2.303
/t) log [A₀]
/[Aₙ]
where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>
[A₀] = 615 calories;
[Aₙ] = 615 - 480 = 135 calories
k = 2.00 * 10⁻²⁰ sec⁻¹
substituting the values in the equation of the rate constant;
2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)
(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)
t = 2.303 / 3.037 * 10⁻²⁰
t = 7.58 * 10¹⁹ seconds