Answer:
Option B, aspirin’s ester group provides greater digestibility to aspirin
Explanation:
Aspirin ester group has three parts
- carboxylic acid functional group (R-COOH)
- ester functional group (R-O-CO-R')
- aromatic group (benzene ring)
Aspirin is a weak acid and hence it cannot dissolve in water readily. The reaction of Aspirin ester group with water is as follows -
aspirin
(acetylsalicylic acid) + water → salicylic acid + acetic acid
(ethanoic acid)
Aspirin passes through the stomach and remains unchanged until it reaches the intestine where it hydrolyses ester to form the active compound.
Answer:
0.2g
Explanation:
Given parameters:
Number of moles of H₂ = 0.4mol
Number of moles of O₂ = 0.15mol
Unknown:
Mass of reactant that would remain = ?
Solution:
To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.
The expression of the reaction is :
2H₂ + O₂ → 2H₂O
2 mole of H₂ will combine with 1 mole of O₂
But given; 0.4 mole of H₂ we will require 
= 0.2mole of O₂
The given number of oxygen gas is 0.15mole and it is the limiting reactant.
Hydrogen gas is in excess;
1 mole of oxygen gas will combine with 2 mole of hydrogen gas
0.15 mole of oxygen gas will require 0.15 x 2 = 0.3mole of hydrogen gas
Now, the excess mole of hydrogen gas = 0.4 mole - 0.3 mole = 0.1mole
Mass of hydrogen gas = number of mole x molar mass
Molar mass of hydrogen gas = 2(1) = 2g/mol
Mass of hydrogen gas = 0.1 x 2 = 0.2g
Answer:
Al2O3 + 3Cl2 + 3 C → 2 AlCl3 + 3 CO
Explanation:
Answer:
#1: 0.00144 mmolHCl/mg Sample
#2: 0.00155 mmolHCl/mg Sample
#3: 0.00153 mmolHCl/mg Sample
Explanation:
A antiacid (weak base) will react with the HCl thus:
Antiacid + HCl → Water + Salt.
In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.
Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:
Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl
Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl
Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl
The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.
Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:
#1: 2.178mmol / 1515mg
#2: 2.253mmol / 1452mg
#3: 2.219mmol / 1443mg
<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
Answer:
a. 27g/mol
b. 1.85 x 10^5 moles
Explanation:Please see attachment for explanation
