Every mole of CH4 used, three moles of H2 are produced, so 2 moles of CH4, would be 6 moles of H2 produced
Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



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
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(According to Hess Law)


Answer:
The correct option is c
Explanation:
The chemical equation for the reaction of Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image
Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers
Answer:
The periodic table has gone through many changes since Dmitri Mendeleev drew up its original design in 1869, yet both the first table and the modern periodic table are important for the same reason: The periodic table organizes elements according to similar properties so you can tell the characteristics of an element just by looking at its location on the table
Explanation:
hope it helps you
D. F
Molecules are a group of bonded atoms but Fluorine stands on its own