A large activation energy is required to break the double bonds of unsaturated hydrocarbons.
<h3>What are Unsaturated hydrocarbons ?</h3>
Unsaturated hydrocarbons are defined as the hydrocarbons in which double or triple bonds are present between two adjacent carbon atoms. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is CnH2n and CnH2n-2
- In unsaturated hydrocarbons, more number of bonds are formed, thus the bond strength of the bonds formed will be more because the orbitals come closer to each other.
- As, bond strength of unsaturated hydrocarbons are more. So, more energy will be required to break the bond between them.
Learn more about Unsaturated hydrocarbons here:
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Fecl3 has a RFM of 56 + (35.5 * 3) = 162.5
so find out its moles 16.5 ÷ 162 5 = aprox. 0.1M
then there is a 1 : 3 ratio so 0.1 * 3 = 0.3
then times that by the avogado constant which is 6.02×10^23
0.3 × 6.02×10^23 = ans
hope that helps
Answer:
Molar Concentration = 
= 
= 13.33
No. of H+ ions present = 13.33
pH value = - log[13.33]
= -1.12
Explanation:
The equivalence point, or stoichiometric point, of a substance response is the point at which synthetically identical amounts of reactants have been blended. As such, the moles of corrosive are equal to the moles of base, as per the condition (this doesn't really infer a 1:1 molar proportion of acid:base, simply that the proportion is equivalent to in the condition). It tends to be found by methods for a marker, for instance phenolphthalein or methyl orange. The endpoint (identified with, however not equivalent to the equivalence point) alludes to the point at which the marker changes shading in a colorimetric titration.
Solutions for your questions are the following:
1. remaining amount is equal to:960 g : 100% = 30 g : xx = 30 g * 100% / 960 g
= 3.125%
= 0.03125
Now, we use this formula to calculate the number of half-lives:(1/2)ⁿ = x,
so,(1/2)ⁿ = 0.03125
to calculate n, use this equation:
n*log(1/2) = log(0.03125) n = log(0.03125)/log(1/2)
= log(0.03125)/log(0.5)
= -1.505/-0.301
n=5
Ifn = 5T = 15 min
Then
L = T/nL = 15 min/5
= 3 minutes
2. You should pick filtration. It is use to separate heterogeneous mixtures just like in the problem stated.
