Answer:
The correct appropriate will be Option 1 (Acid anhydrides are less stable than esters so the equilibrium favors the ester product.)
Explanation:
- Acid anhydride, instead of just a carboxyl group, is typically favored for esterification. The predominant theory would be that Anhydride acid is somewhat more volatile than acid. This is favored equilibrium changes more toward the right of the whole ester structure.
- Extremely responsive than carboxylic acid become acid anhydride as well as acyl chloride. Thus, for esterification, individuals were most favored.
The other options offered are not relevant to something like the scenario presented. So, the solution here is just the right one.
Remember that:
number of moles = mass/molar mass
First, we get the molar mass of the nitrogen gas molecule:
It is known the the nitrogen gas is composed of two nitrogen atoms, each with molar mass 14 gm (from the periodic table)
Therefore, molar mass of nitrogen gas = 14 x 2 = 28 gm
Second we calculate the mass of the precipitate:
we have number of moles = 0.03 moles (given)
and molar mass = 28 gm (calculated)
Using the equation mentioned before,
mass = number of moles x molar mass = 0.03 x 28 = 0.84 gm
Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
You need to find the whole molar mass of the compound using the periodic table to add the values.
Na2CO3= (2 x 23.0) + 12.0 + (3 x 16.0)= 106 g/mol
H2O= 10 x [ (2 x 1.01 ) + (16.0) ]= 180.2 g/mol
the total molar mass is 106 + 180.2 = 286.2 g/mol
the percentage of water you can find by doing "parts over the whole"
H2O%= 180.2 / 286.2 X 100= 63.0%
<span>31.9 grams butane needed to produce 96.7 grams CO2
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