In organic chemistry, there are already known reaction for synthesis that have been studied and experimented by scientists. For reactions involving a Nickel catalyst in the presence of hydrogen gas, the reaction would be hydrogenation of alkene to yield an alcohol. These reaction is anti-Markovnikov, or proceeds by violating the rule of Markovnikov. His rule states that 'the rich gets richer'. This means that the heavily substituted C atom of the alkene chain would receive another substituent. Therefore, the complete reaction would be
4-ethyl-3-hexene -------> 4-ethyl-3-hexanol
In structural formula, the reaction goes as follows as shown in the picture.
To solve this problem, we need to set-up algebraic expression. First, use variables to represent the number of beads.
Let: x = number of Xavier's beads
y = number of Y<span>aozhou's beads
z = number of Zara's beads
</span>It is important to note that since we have three unknowns, we should also have three independent equations. <span>Based on the given statements, we have the following three equations:
(1) x + y = 438
(2) x + z = 204
(3) y = 3z
Substitute y in terms of z in equation (1). Then multiply equation (2) with -1.
x + 3z = 438
-(x + z = 204)
-------------------
2z = 234
z = 117
From equation (3),
y = 3(117) = 351
Using equation (1),
x = 438 - 351 = 87
Thus, Xavier had 87 beads.</span>
HI
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Answer: Yes, a given amount of Hydrogen would react with different masses of the two isotopes of chlorine, and no, this does <em>not</em> conflict with the Law of Definite proportions
Explanation:
About 76% percent of Cl is found in the Cl-35 isotope, and about 24% in the Cl.37 isotope. that means that about 24% of Cl nuclei have 2 more neutrons than the average Cl nucleus.
So, if 
reacts with 
, 76% of the Hydrogen that reacted will react with Cl-35, and the rest will react with Cl-37. Why does this not conflict with the law of definite proportions? Because each Hydrogen atom ends up paired to a single Chlorine atom! Moreover, the proportion of Cl-35 to Cl-37 remains constant in all samples of Chlorine that are naturally found, thus we will always find the same proportion of Chlorine to Hydrogen in any HCl sample we come across. Thus the weight of a mol of Cl will always be 
or 35.45 if we had done this calculation with more significant digits.
Therefore 1 mol of (2 grams) will always react with 1 mol of (35.45g*2=70.9 grams), and this is a definite proportion.
