Answer: a = 2 ; f = 5 ; b = 2 ; g = 2 ; c = 2 ; h = 2 ; d = 4 ; i = 5 ; e = 3 ; j = 7
Explanation: Some rules to follow while calculating sig figs is
1. If a number like 4500 is present, only two sig figs are counted, but none of the zeros are, but if 4500. has a decimal point present, then you should count all the numbers available.
2. If a number like .0005 is present, only count 5 as a sig fig, however if the number is .00050, count the 0 after the 5 in this example (this would then have two sig figs.
<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
Answer:
Such a waste of points but lol
Answer:
Theoretical moles of V are 1.6 moles
Explanation:
The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.
In the reaction:
V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)
Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:
4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>
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I hope it helps!
Answer:
The answer to your question is letter B, 2-methylhexane.
Explanation:
Remember that for naming organic compounds first, we need to look for the largest chain of carbons.
In your example, the largest chain is horizontal and has 6 carbons.
Later, we need to circle all the branches, in your example there is only one branch located close to the left side
After that, we number the carbons of the main chain, starting in the corner with more branches, in your example we start from the first carbon on the left.
Finally, start naming the number of the carbon branch, later hte name of the branch and finally the name of the main chain.
