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1. The dosage for the 165 pound adult would be 8.66 mg/kg.

10 grains x 1 gram x 1000 mg x 1 pound x 1000 grams = 10,000,000 mg

165 lbs 15.43 grains 1 gram 453.6 g 1 kg 1,134,842.92 kg

= 8.66 mg

1 kg

2. Part A: Applying the law of simplicity to compound I can be determined after I have some addition information; specifically, the mass % of Z. The ratio of the atomic weight of X to Z is listed in the fifth column of the table below.

Part B: The simplest formula for each of compounds II to IV is listed in the fifth column of the table below. This is the best I can do because the law of simplicity is an essential presumption that one of the compounds has a ratio of 1:1 and we were given the information that compound I would be that . That assumption is the basis for the calculations that I did.

|Compound |Mass % X |Mass % Z |gX : 1gZ |Ratio |

|I = X1Z1 |61.37 |38.63 |1.589 g |1:1 |

|II = X2Z3 |51.44 |48.56 |1.059 g |2:3 |

|III = X1Z2 |44.27 |55.73 |0.794 g |1:2 |

|IV= X2Z5 |38.86 |61.14 |0.636 g |2:5 |

gX = p * AWx Compound I: 1.589 g = AWx

gZ q AWz 1 g AWz

Compound II: 1.059 g = x * 1.589 g X

1 g z 1 g Z

1.059 g = x

1.589 g z

.66 = 2/3

Compound III: 0.794 g = x * 1.589 g X

1 g z 1 g Z

0.794 g = x

1.589 g z

.499 = ½

Compound IV: 0.636 g = x * 1.589 g X

1 g z 1 g Z

1.059 g = x

1.589 g z

.4 = 2/5

Part C: Using the previous information that was based on a presumption, we can determine that the two elements in the compounds are Vanadium (symbol V) and Sulfur (symbol S). We did this by cross multiplying the proportion.

AWx = 1 = x ( 1.589 = x ( x = 50.848 ( 51 on the modern periodic table

AWz 32 S 1 32

So 32 = Sulfur (S) and 51 = Vanadium (V)

Part D: Knowing the chemical symbols for x and z does allow me to write compounds using the empirical data, but this data was based on the presumption of the law of simplicity. In fact, the atomic weights that generated the knowledge that x was Vanadium and z was Sulfur is also hypothetical based on this same presumption.

Compound I ( VS

Compound II ( V2S3

Compound III ( VS2

Compound IV ( V2S5

3. How many atoms of Carbon will weigh 12.0 grams is the question.

From the data and explanation given, we know that an atomic mass unit (amu) = 1.661 x 10-24grams and 12C = 12.0 amu. These are references or definitions of which we need to be aware in order to determine how many atoms equal one mole of Carbon.

12 g C * 1 amu * 1 atom C = 6.02 x 1023 atoms = 1 mole

1 1.661 x 10-24g 12.0 amu

This formula can be used for any element and compound in order to determine the amount of atoms or molecules present.

4. Based on the information gleaned from the formula in the previous question, we can now move beyond where Dalton was.

Part A: The mass (in g) of each compound in a sample of the compounds are:

H2O = 2400 g ( m = D * V ( m = 1.00 g/mL x 2400 mL

C8H18 = 1687 g ( m = D * V ( m = .703 g/mL x 2400 mL

N2O = 4.32 g ( m = D * V ( m = 1.80 g/L x 2.4L

NO2 = 4.51 g ( m = D * V ( m = 1.88 g/L x 2.4L

Part B: The number of molecules present in 2.4 L samples would be:

Water = 8.03 x 1025 molecules

Octane = 8.91 x 1024 molecules

Nitrous Oxide = 5.91 x 1022 molecules

Nitrogen Dioxide = 5.90 x 1022 molecules

In order to calculate the number of molecules present, I used the formula:

1 amu x mass in grams x 1 molecule = molecules = molecules

1.661 x 10-24 grams 1 amu/compound moles

Part C: The mass percents of each element in each compound are found through the ratio of the AW(#atoms)/total atoms in the compound. Thus:

Water = 11 % hydrogen

89 % oxygen

Octane = 84% carbon

17% hydrogen

Nitrous Oxide = 63% nitrogen

36% oxygen

Nitrogen Dioxide = 30% nitrogen

70 % oxygen

Please note: I will send the work for this part in an hour or so. My dinner is burning.

Part D: If I apply the Law of Multiple Proportions to the nitrous oxide/nitrogen dioxide pair, I can write the number of grams of nitrogen per gram of oxygen as a ratio of whole numbers. There would be four grams of nitrogen per one gram of oxygen.

63 : 30_ ( 1.75 : 43 ( 4:1

36 70

5. Part A: Density of the nucleus would be: 3.17 x 1015 g/cc.

Density of the atom would be: 3.17 g/cc.

If the nucleus of this atom were scaled up to the size of a garden pea, the mass would be 4.69 x 108 tons.

See above note.

Part B: I don’t even know in which direction I should head with this one.

6. If the chemical reaction was 3[pic]and 2 equals 4 then the final reaction would be 20 triangles + 1 square.

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