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zhuklara [117]
3 years ago
8

there is a total of seventeen constitutional isomers for the molecular formula c5h13n. draw the skeletal formula of all three co

nstitutional isomers that are tertiary amines.

Chemistry
1 answer:
MAXImum [283]3 years ago
5 0
Amines are derivatives of Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;

Primary Amines:
                          In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.

Secondary Amines:
                             In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.

Tertiary Amines:
                         In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.

Below are three isomers of tertiary amines with molecular formula C₅H₁₃N.

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Given 5.0 grams of lead (II) nitrate and 3.0
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Answer:

                     1.70 g of NaNO₃

Explanation:

                     The balance chemical equation for given double displacement reaction is,

                               Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO₃

Step 1: <u>Calculate moles of each reactant:</u>

                                   Moles =  Mass / M.Mass

For Pb(NO₃)₂:

                                   Moles =  5.0 g / 331.21 g/mol

                                   Moles =  0.0150 moles

For NaI:

                                   Moles =  3.0 g / 149.89 g/mol

                                   Moles =  0.020 moles

Step 2: <u>Calculate Limiting reagent as;</u>

According to equation,

                     1 mole of Pb(NO₃)₂ reacts with  =  2 moles of NaI

So,

              0.0150 moles of Pb(NO₃)₂ will react with  =  X moles of NaI

Solving for X,

                     X =  2 mol × 0.0150 mol / 1 mol

                     X =  0.030 moles of NaI

Hence, it means that NaI is the limiting reagent therefore, it will control the yield of Sodium Nitrate.

Step 3: <u>Find out moles of NaNO₃ formed:</u>

According to equation,

                     2 mole of NaI produced =  2 moles of NaNO₃

So,

              0.020 moles of NaI will produce   =  X moles of NaNO₃

Solving for X,

                     X =  2 mol × 0.020 mol / 2 mol

                     X =  0.020 moles of NaNO₃

Step 4: <u>Calculate Mass of NaNO₃ produced;</u>

As,                Moles  =  Mass / M.Mass

Or,

                    Mass  =  Moles × M.Mass

Putting values,

                    Mass  =  0.020 mol × 84.99 g/mol

                    Mass  =  1.70 g of NaNO₃

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Answer:

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Explanation:

your welcome

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