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How to Solve Problems and Consider Significant Figures Special Issue
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Every problem has its own “anatomy”. A problem consists of tale and data . Tale may be boring and funny. Data may be necessary and unnecessary.
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The tale and the data : Silver (Ag: Z = 47 ) has 46 known isotopes, but only two occur naturally , 107 Ag and 109 Ag . Given the following mass spectrometric data, calculate the atomic mass of silver: 107 Ag: atomic mass 106.90509 , abundance 51.84 109 Ag: atomic mass 108.90476 , abundance 48.16
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Let us handle a problem step by step: Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral.
Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral.
Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S . Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral .
m 1 (Zn)=327.1 g ; m 1 (S)=160.4 g ; m(ore)=541 kg
Sphalerite is a mineral, a chief ore of zinc. A sample of sphalerite contains 327.1 g of Zn and 160.4 g of S. Please calculate masses (g) of zinc and sulfur in 541 kg of this mineral .
To find mass of zinc and sulfur in a given mass of their compound, we must know their mass fractions (or mass percents) in this compound.
To find the mass fraction of zinc, we should divide mass of zinc in the sample by the mass of this sample. The same about sulfur.
To find the mass of the sample, we have to add masses of zinc and sulfur in this sample.
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5. Solve (slide 1) starting from the end of the speculation:
Find the mass of the ore sample: 327.1g + 160.4g = 487.5g
Find the mass fraction of zinc: 327.1g/487.5g = 0.67097436
Consider the number of significant figures: 3 .
Round the value of zinc mass fraction: 0.67097436 ≈ 0.671
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5. Solve (slide 2) starting from the end of the speculation:
Find the mass fraction of sulfur: 160.4g/487.5g = 0.32902564
Consider the number of significant figures: 3 .
Round the value of sulfur mass fraction: 0.32902564 ≈ 0.329
or just 1.000 - 0.671 = 0.329 - much faster!
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5. Solve (slide 3) starting from the end of the speculation:
Find the mass of zinc in 541 kg of the mineral: 0.671 x 541 kg = 363.011 kg
Consider the number of significant figures: 3.
Round the mass of zinc: 363.011kg≈363 kg
Convert to grams: 363kg=363000g=3.63x10 5 g
Finalize for WebAssign: 3.63e5
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5. Solve (slide 4) starting from the end of the speculation:
Find the mass of sulfur in 541 kg of the mineral: 0.329x541kg=177.989kg
Consider the number of significant figures: 3.
Round the mass of sulfur: 177.989kg≈178kg
Or just 541 kg - 363 kg = 178 kg - much faster!
Convert to grams: 178kg=178000g=1.78x10 5 g
Finalize for WebAssign: 3.63e5
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Let us handle another problem step by step: Gallium has two naturally occurring isotopes, 69 Ga (isotopic mass 68.9256 amu, abundance 60.11%) and 71 Ga (isotopic mass 70.9247 amu, abundance 39.89%). Please calculate the atomic mass of gallium.
Gallium has two naturally occurring isotopes, 69 Ga (isotopic mass 68.9256 amu, abundance 60.11%) and 71 Ga (isotopic mass 70.9247 amu, abundance 39.89%). Please calculate the atomic mass of gallium.
Gallium has two naturally occurring isotopes, 69 Ga (isotopic mass 68.9256 amu, abundance 60.11%) and 71 Ga (isotopic mass 70.9247 amu, abundance 39.89%). Please calculate the atomic mass of gallium .
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