c 2.998 3 10 T0 h =6.626 x 10-34 Jsec R = 0.0821 LatmimoIK R 8.314 J/molK = 8.31
ID: 697699 • Letter: C
Question
c 2.998 3 10 T0 h =6.626 x 10-34 Jsec R = 0.0821 LatmimoIK R 8.314 J/molK = 8.314 kgm2/sec2moIK 1. There are three isotopes of the element magnesium, 24Mg(23.98504 amu), 25Mg(24.98584 amu), 28Mg(25.98259 amu). Calculate the average atomic mass of magnesium knowing that the percent abundance of 24Mg is 78.99% & 25Mg is 10.00%. (10 points) 2. 15.00 mL of 0.1008 M potassium hydroxide aqueous solution are needed to neutralize a 0.1544 g sample of a diprotic acid. What is the molar mass of the unknown diprotic acid? An elemental analysis of the diprotic acid indicates that it is composed of 5.89% H, 70.6% C, and 23.5% ) by mass. What is its molecular formula? (16 points)Explanation / Answer
average atomic mass = isotope1 *percent abundance + isotope 2*percent abundance + isotope3 + percent abundance
= 23.98504*0.7899+24.98584*0.1+ 25.98259*0.1101
= 24.3amu >>>>answer
2.H2A + 2KOH -------------> K2A + 2H2O
no of moles of KOH = molarity * volume in L
= 0.1008*0.015 = 0.001512moles
from balanced equation
2 moles of KOH react with 1 mole of H2A
0.001512 moles of KOH react with = 1*0.001512/2 = 0.000756 moles of H2A
molar mass of H2A = mass of H2A/ no of moles
= 0.1544/0.000756 = 204.23g/mole
element % A.Wt relative number simple ratio
C 70.6 12 70.6/12 = 5.89 5.89/1.45 = 4
H 5.89 1 5.89/1 = 5.89 5.89/1.45 = 4
O 23.5 16 23.5/16 = 1.45 1.45/1.45 = 1
The empirical formula = C4H4O
Empirical formula weight = 12*4 + 1*4 + 16 = 68g/mole
molecular formula = (empirical formula)n
n = 204.23/68 = 3
molecular formula = (C4H4O)3
= C12H12O3 >>>>answer
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