32 3.0 liters of helium at 5.6 atm pressure and 25 C and 4.5 liters of neon at 3
ID: 697592 • Letter: 3
Question
32 3.0 liters of helium at 5.6 atm pressure and 25 C and 4.5 liters of neon at 3.6 atm and 25° C are combined at constant temperature in a 9.0 liter flask. What is the total pressure (in atm) in the 9.0 liter flask? A. 1.0 B. 1.5 C. 2.6 D. 3.7 E. 9.2 33 31.6 mL of 0.1985 M sodium hydroxide solution is needed to neutralize 20.0 mL of an acetic acid solution (density = 1.025 g/mL), by the following reaction: H3CCOOH + NaOH H3CCOONa + H2O What is the weight percent of acetic acid (H CCOOH) in the original acetic acid solution? A. 0.314% B. 2.51% C. 0.918% u, il D. 1.84% E. 3.68%Explanation / Answer
32) V1 = 3.0L ; P1 = 5.6 atm
V2 = 4.5 L; P2 = 3.6 atm
Vtotal = 9.0 L ; Pfinal = ?
From Boyles law;
P1V1 + P2V2 = Vtotal.Pfinal
5.6 atm x 3.0 L + 3.6 atm x 4.5 L = 9.0 L x Pfinal
Therefore, Pfinal = 33/9 = 3.67 atm
Hence, the answer is: D (3.7 atm)
33) balanced reaction is :
H3CCOOH + NaOH ---> H3CCOONa + H2O
Therefore, 1 mol of acetic acid reacts with 1 mol of NaOH.
No. of moles of NaOH used for neutralization = molarity x volume in L
= 0.1985 mol/L x 0.0316 L
= 0.0062726 mol
Therefore, in the gibe 20 mL of acetic acid solution, we have 0.0062726 mol of acetic acid.
Mass of acetic acid = no. moles x mol.wt
= 0.0062726 mol x 60.05 g·mol1
= 0.377 g
Total mass of the solution = Volume x density of the solution
= 20 mL x 1.025 g/mL
= 20.5 g
Therefore % of acetic acid = (mass of acetic acid/ total mass of solution)x100
= (0.377 g/20.5 g)x100
= 1.839
Hence the answer is: D (1.84%)
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