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Dalton Oxygen Research Paper

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Dalton Oxygen Research Paper
Chapter 2

22. A student heats 0.5585 g of iron with 0.3550 g of sulfur. She reports that she obtains 0.8792 g of iron sulfide and recovers 0.0433 g of unreacted sulfur. Show by calculation whether or not her results obey the law of conservation of mass.
Total mass initial = 0.5585 g + 0.3550 g = 0.9135 g
Total mass final = 0.8792 g + 0.0433 g = 0.9225 g
These two values should be equal by the law of conservation of mass. These results do not obey the law of conservation of mass. Possibly she made a mistake recording her numbers.

28. When 31 g of phosphorus reacts with oxygen, 71 g of an oxide of phosphorus is the product. What mass of oxygen is needed to produce 13 g of this product?
31g + x = 71g
71g - 31g = 40g oxygen
This means that in
…show more content…
Compare Dalton’s model of the atom with the nuclear model of the atom.
Dalton atomic model is the simplest model. Atoms only described as a very small ball but massive. This model is still not familiar with the charge within the atom.
The nuclear model the atoms are not massive but hollow objects. Electrons are not scattered evenly in the nucleus but are at great distances from the nucleus. Rutherford's theory also states that the electrons are not silent but orbit around the nucleus.

9. What are the symbol, name, and atomic mass of the element that has 18 protons in the nucleus of its atoms?
Symbol= Ar (Argon) mass= 39.948g

10. Explain what is meant by the term atomic mass.
Atomic mass is the weight of protons and neutrons combined in an element

14. How many electrons are there in the neutral atoms of the elements listed in Question 13? Helium= 2 Oxygen= 8 Sodium=11 Magnesium= 12 Chlorine=17 Sulfur= 16

16. Fill in the following table: Element Mass Number Protons Neutrons
- Cobalt 60 27 32
- Fluorine 19 9 10
- Uranium 238 92 143
- Lead 207 82

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