In a chemical change, the atoms in the reactants rearrange themselves and bond together differently to form one or more new products with different characteristics than the reactants. When a new substance is formed, the change is called a chemical change.
the refractive index of the liquid is 1.476
The refractive index, which has no dimensions, measures how quickly light passes through a substance.
It can also be described as the difference between the speed of light in a vacuum and a medium.
Refractive index is equal to the product of the light's liquid and vacuum speeds.
Therefore.
speed of light in vacuum = 4.96 km/t
speed of light in liquid = 3.36 km/t
Refractive index = 4.96/3.36
Refractive index =1.476
Therefore, the refractive index of the liquid is 1.476
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Answer:
See Explanation and image attached
Explanation:
Methane is an alkane. The commonest chemical reaction that alkanes undergo is substitution. During a substitution reaction, one or more atoms of hydrogen is/are replaced in the alkane.
In methane, in the presence of sunlight and molecular chlorine gas, a homolytic fission of Cl2 occurs to yield chlorine radicals in an initiation step.
The propagation steps involve reaction of the methane with chlorine radicals. Certain intermediates continue to be formed along the way until the tetrachlorination product is finally obtained.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
