Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Answer:
you are not posted question
Answer:
the first one
Explanation:
as it contains in the product N so its the forst
<span>Cracking of hydrocarbons involves thermal decomposition.
This means that large hydrocarbon molecules break into
smaller molecules when they are heated. The <span>hydrocarbons
</span>are boiled and the <span>hydrocarbon gases </span>are either
mixed with steam and heated to a very high temperature or
passed over a hot <span>powdered </span>aluminium oxide catalyst.
The catalyst works by providing the <span>hydrocarbon gases
</span>with a convenient surface for the cracking to take place.</span>
<span>For example, decane (an alkane with 10 carbons)
can be cracked to produce octane and ethene.</span>
decane<span> octane + ethene.
<span>C10H22</span>(g) <span>C8H18</span>(g) + C2<span>H4</span>(g)</span>
<span>Octane is used as petrol.
Ethene is used in the manufacture of polymers.</span>
<span>Cracking an alkane produces a smaller alkane plus an alkene.
If you add up the number of hydrogen atoms in the
above reaction, you will see that there are 22 on each side.
An alkene is produced because the original alkane does
<span>not </span>have enough hydrogen atoms to produce two more<span> alkanes</span>.</span>
<span>
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<span>hope this helps </span>
