Chemistry Hw Review | Framingham State University

University:
Framingham State University
Course:
CHEM 101 | The Chemistry of Life
Academic year:

2020
Views:

288

Pages:

2
Author:

Lacey

Average Atomic Mass calculation: ● (Mx = A1 M1 + A2 M2 + A3 M3 + (blank) ) ● Where mx is the average atomic mass of element x. ● Which has isotopes with masses m1, m2, m3, “x “. ● The natural abundances of which are expressed in decimal form. A1, a2, a3, “x“ Analyze ● I have to compare the two isotopes and their weight. ● Lighter isotope is more abundant than the average atomic mass will be less than the average if both isotopes were equally abundant. ● If boron -10 and boron -11 average mass would be 10.5. Since the average is closer to 11 than it is to 10. Heavier isotopes are more abundant. Mole ● Cobalt (II) iodine hexahydrate (col2 - 6H2O) CoI2 x 6H2O 1 Co + 2 I + 12 H + 6 Oxy 1 (58.93) + 2 (120.90)+ 12 (1.008) 6(15.99) 58.93 + 253.8 + 12.096 + 12.096 + 95.94 = 420.8 amu ● How many moles in 400g of carbon? - C = 12.001 400 / 12.001 = 33.0 g/mol ● Determines the moles of Na3AlF6 provided by 3.25 1000g = kg Na3AlO2 : Na3AlF6 3:1 Determine the kilograms of NaAlO2 needed to produce 15.5 mol Na3AlF6 3810kg is the same as 3.81 kg NaAlo2

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