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stepan [7]
1 year ago
15

Which ketone in each pair is more reactive?

Chemistry
1 answer:
KiRa [710]1 year ago
4 0

Answer:

a. 2-heptanone is more reactive than 4-heptanone

b. chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone

Explanation:

The reactivity of the carbonyl compound (ketone ) is affected by the steric effect. The steric effect is a hindrance that occurs in the structure or reactivity of a molecule, which is affected by the physical size and the proximity of the adjacent parts of the molecule.

Between 2-heptanone or 4-heptanone, 2-heptanone is more reactive than 4-heptanone. This is because 2-heptanone is less affected by the steric hindrance, unlike the 4-heptanone.

Similarly, the reactivity of the carbonyl compound (ketone) is also affected by the polarity on the carbon compound, which is associated with how electronegative the substituent attached is to the carbonyl compound. From the periodic table, the electronegativity of the Halogen family decreases down the group. Therefore chlorine is more electronegative than bromine.

As such, chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone.

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How many acidic protons are there in 0.6137 g of KHP?
andreyandreev [35.5K]

0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.6137 g

Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now,  1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So;  0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

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1 year ago
100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

1

2

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

1

1

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

1

4

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

6 0
2 years ago
Read 2 more answers
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Anna11 [10]
Molar mass  CaCl₂ =  110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 =  0.7686 / V

V = 0.7686 / 0.788

V = 0.975 L

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