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2 years ago

How much aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate? Answer in units of mol.

Chemistry

1 answer:

Bond [772]2 years ago

7 0

0.77 moles of aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate.

<h3>Equation of reaction</h3>

Al + Pb(NO₃)₂ ---> Al(NO₃)₃ + Pb (s)

From the equation of reaction:

1 mole of aluminum reacts with 1 mole of lead nitrate to produce 1 mole of aluminum nitrate

1 mole aluminum has a mass of 27 g

Number of moles of aluminum in 21 g = 21/27 =0.77 moles

0.77 moles of Aluminum produces 0.77 moles of aluminum nitrate

Therefore, 0.77 moles of aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate.

Learn more moles at: brainly.com/question/13314627

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Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the product

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Answer:

See figure 1

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Now the question is: ¿These compounds can react by a nucleophile acyl substitution reaction? in other words ¿These compounds can produce an amide? 

Due to the nature of the compounds (base and acid), the nucleophile (methylamine) doesn't have the ability to attack the carbon of the carbonyl group due to his basicity. The methylamine will react with the acid-producing a positive charge on the nitrogen and with this charge, the methylamine loses all his nucleophilicity.

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3 years ago

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2 years ago

A 0.15 M NaOH solution has a volume of 0.125 L but is then diluted to 0.15 L. What is the concentration of the new solution

tresset_1 [31]

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1 year ago

an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with

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Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.

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3 years ago