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4 years ago

Which series is the most reactive series? alkyne alkene alkane methane

1 answer:

6 0

Hello.

Alkali metals are the most reactive group in the periodic table, not to be confused with Alkaline earth metals.

Also, Methane is a gas, not a group on the periodic table.

they contain triple bond, one sigma bond and 2 pi bonds. Pi bonds are weak and easily broken making the compound active.

Chemical reaction is the process of breaking down bonds in reactants and forming new bonds in products. 
In Alkynes breaking down bonds is easier so they are more active.

Have a nice day

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A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N

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3 years ago

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4 years ago

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A) A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M

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Answer:

i) 5.00 mL of 1.00 M NaOH: before the equivalence point

ii) 50.0 mL of 1.00 M NaOH: before the equivalence point

iii) 100 mL of 1.00 M NaOH: at the equivalence point

iv) 150 mL of 1.00 M NaOH: after the equivalence point

v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

$M=n/v\implies n=M\times V=1.00M\times 0.100L=0.100mol$

2. Now calculate the number of moles of NaOH for every condition (addition)

i) 5.00 mL of 1.00 M NaOH

$n=0.00500liter\times 1.00M=0.00500molNaOH$

Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

ii) 50.0 mL of 1.00 M NaOH

$n=0.0500liter\times 1.00M=0.0500molNaOH$

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

iii) 100 mL of 1.00 M NaOH

$n=0.100liter\times 1.00M=0.100molNaOH$

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

iv) 150 mL of 1.00 M NaOH

$n=0.150liter\times 1.00M=0.150molNaOH$

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

v) 200 mL of 1.00 M NaOH

This is more volume of NaOH, then this is also after the equivalence point.

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