Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
Answer:
0.020 moles of 
can be formed
Explanation:
1. First determine the number of moles of LiOH.
Molarity is given by the following expression:
Solving for moles of solute:
moles of solute = M * Liters of solution
Converting 175.0mL to L:
Replacing values:
moles of solute = 0.227M*0.175L
moles of solute = 0.040
Therefore there are 0.040 moles of LiOH
2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of produced:
As the problem says that there are excess of 
, the limiting reagent is the LiOH.
can be formed
Answer:
D
Explanation:
The scientist identifies the need to make a drug to cure the respiratory illnesses.
Answer:
B)
Explanation:
It is the theme of the passage.
Answer:
20 kJ/mol
Explanation:
From ∆G°= -RTlnK
But
Ag2SO4(s)<----------->2Ag+(aq) + SO4^2-(aq)
Ksp= [2Ag+]^2 [SO4^2-]
But Ag+ = 0.032M
Ksp= (2×0.032)^2 (0.032)
Ksp= 1.31072×10^-4
∆G°= -RTlnK
∆G°= -(8.314× 298×(-8.93976))= 20KJmol-1( to the nearest KJ)
