Main advantages of DDT are the insecticide property and the knoledge about its chemical synthesis.
Explanation:
At that time in 1939 it was discovered that DDT have insecticide properties. It is a useful property because it allows inhibition of insects populations in large areas. Killing insects will reduce the diseases transmitted by them as typhus and malaria. More over you prevent the destruction of the agricultural crops by the harmful insects.
However the synthesis of the molecule was known way back in 1874. From that time it was plenty of time in which chemistry knowledge evolved so the synthesis at kilograms scale was implemented. High quantities of DDT molecule become available for the market so that in 1945 was available as agricultural insecticide.
It was discovered that DDT have bad effects for human health and also over time some insects developed resistance and their were not affected anymore by the molecule.
You may find the chemical structure of DDT in the attached figure.
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Answer:
Explanation:
In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.
We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.
The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.
To ascertain the oxidation state, we have to comply with some rules:
- The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
- The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion.
For example, let us find the oxidation state of Cr in Cr₂O₇²⁻
This would be: 2x + 7(-2) = -2
x = +6
We see that the oxidation number of Cr, a transition metal in the given ion is +6.
Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
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