Your SlideShare is downloading. ×
Problem Solving 1
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Saving this for later?

Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime - even offline.

Text the download link to your phone

Standard text messaging rates apply

Problem Solving 1

2,625
views

Published on

Published in: Technology, Spiritual

0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
2,625
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
122
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. How to Solve Problems and Consider Significant Figures Special Issue
  • 2. 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.
  • 3. 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
  • 4. 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.
  • 5. 1.Read the problem carefully:
    • 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.
  • 6. 2. Flesh out the given data:
    • 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
  • 7. 3. Understand what is required:
    • 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 2 (Zn)= ? ; m 2 (S)= ?
  • 8. 4.Speculate starting from the question posed:
    • 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.
  • 9. 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
  • 10. 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!
  • 11. 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
  • 12. 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
  • 13. 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.
  • 14. 1.Read the problem carefully:
    • 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.
  • 15. 3. Understand what is required:
    • 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 .
  • 16. 4.Speculate starting from the question posed:
    • To find the atomic mass of gallium, we must add the portions contributed by each isotope of gallium.
    • To find a portion contributed, we have to multiply an isotope’s mass number by its fractional abundance.
  • 17. 5. Solve (slide 1) starting from the end of the speculation:
    • Find the portion contributed by 69 Ga: 68.9256x60.11% / 100%=41.431178 or 68.9256x0.6011=41.431178
    • Consider the number of significant figures: 4 .
    • Round the value of 69 Ga portion: 41.431178 ≈ 41.43
  • 18. 5. Solve (slide 2) starting from the end of the speculation:
    • Find the portion contributed by 71 Ga: 70.9247x39.89% / 100%=28.291863 or 70.9247x0.3989=28.291863
    • Consider the number of significant figures: 4 .
    • Round the value of 71 Ga portion: 28.291863 ≈ 28.29
  • 19. 5. Solve (slide 3) starting from the end of the speculation:
    • Add the portions contributed by 69 Ga and 71 Ga: 41.43+28.29=69.72 . It is atomic mass of gallium.
  • 20. THE END