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.
9 grams of hydrogen gas (H2) will SC Johnson need to react in order to make 1 bottle of Windex.
Explanation:
Balance equation for the formation of ammonia from H2 gas.
N2 + 3H2 ⇒ 2 
Given
mass of ammonia in 1 bottle of windex = 51 gram
atomic mass of ammonia 17.01 gram/mole
number of moles = 
number of moles = 
= 3 moles of ammonia is formed.
in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.
From the equation it can be found that:
3 moles of hydrogen reacted to form 2 moles of ammonia
so, x moles of hydrogen will react to form 3 moles of ammonia.
= 
x = 4.5 moles of hydrogen will be required.
to convert moles into gram formula used:
mass = atomic mass x number of moles (atomic mass of H2 is 2grams/mole)
= 2 x 4.5
= 9 grams of hydrogen.
Answer: Aluminum, Antimony, Arsenic, Bismuth, Carbon, Cadmium, Chromium, Cobalt, etc.
Answer:
Proton is a sub atomic particle while oxygen is an atom ....
Toichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag ∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.