a) The observations I made from the results of the chemist are that every sample of water into the hydrogen and oxygen increases by .5 grams of hydrogen and 4 grams of oxygen.
b) The observation I made from the results of the chemist adding several samples of carbon dioxide into carbon and oxygen is that each sample size increase by .5 grams of carbon and 1.3 grams of oxygen.
c) A law from my observations I could formulate is when H20 is added to hydrogen and oxygen, .5 grams of hydrogen comes from each hydrogen in the H20 and 4 grams of oxygen comes from the oxygen in H20. A law from my observations I could formulate when CO2 is added into carbon is that .5 grams of carbon will come from the carbon from the C02. When C02 is added to oxygen 1.3 grams of oxygen and .1 grams every sample will be added to the oxygen sample.
d) A theory I can determine from the laws is that the beginning element of a compound such as H in H20 and C in C02 will always add the same number of grams to its same element. Such as H20 will always add .5 grams to hydrogen and C02 will always add .5 grams to carbon
Exercise 120 in Ch.2
Answer: The difference in size between the two diameters is 5.6 x 109m2
Work: 2003: 1.04 X 107 mi2 2002: 6.9 X 106mi2
1.04 X 107 mi2 -6.9 X 106mi2 = 10,400,000-6,900,000= 3.5 X 106mi2
3.5 X 106 mi2 X 1600m2/1mi2=5.6 x 109m2
Exercise 112 in Ch.3
We calculate temperature change due to heat transfer with this equation:
Q = m x c x delta T
Where Q stands for the heat in joules, m is the mass of the substance in grams, c is the specific heat capacity in joules per gram per degree Celsius, and delta T is the change in temperature in Celsius. (The symbol Δ means the change in, so ΔT means the change in temperature. Δ also means delta).
In this case, c = 4.186 J/g C (joules per gram per degree C)
Delta T = 1
We need to calculate the mass of the oceans in grams for m
1 Km = 1 x 10^5 cm; we cube this to get:
1 Km^3 = 1 x 10^15 cm^3
Then we can do the conversion -
137 x 10^7 Km^3 x 1 x 10^15 cm^3/1 Km^3 = 1.37x10^24
Now, back to Q = m X c X delta T
Q = 1.37x10^24 x 4.186 x 1
Q = 5.735x10^24 Joules
Exercise 120 in Ch.4
Neutron stars are believed to be composed of solid nuclear matter, primarily neutrons.
a. If the radius of a neutron is 1.0 × 10–13 cm, calculate the density of a neutron in g/cm3. (volume of a sphere = V=4/3πr3)
b. Assuming that a neutron star has the same density as a neutron, calculate the mass in kilograms of a small piece of a neutron star the size of a spherical pebble with a radius of 0.10 mm.
Density = mass / volume
D = 1g / (6.02 x 10^23 x (4/3) x pi x (1.0 x 10^-13)^3)
D = 1g / (6.02 x 10^23 x (4/3) x pi x 10^-39cm^3)
D = 1 / (6.02 (4/3) x pi x 10^-16) g/cm^3
D = 10^16 / (6.02 x (4/3) x pi)
D = 3 x 10^16 / (24.08 x pi) g/cm^3
D = 0.0396565 x 10^16 g/cm^3
Density = 3.966 x 10^14 g/cm^3
Part b: r = 0.10 x 10^-1 cm r = 1.0 x 10^-2 cm
V = (4/3)x pi x (1.0 x 10^-2)^3
V= (4/3) x pi x (1 x 10^-6)
V = 4.188 x 10^-6 cm^3
Mass = V x D
Mass = (4.188 x 10^-6 cm^3) x (3.966 x 10^14 g/cm^3)
Mass = 16609.608 x 10^8 g
Mass = 1.661 x 10^12 g
Mass = 1.661 x 10^9 kg
Exercise 100 in Chapter 5
The chemical formula of hemoglobin is C2952H4664O832N812S8Fe4. Calculate the formula mass of hemoglobin.
C (Carbon) 2952 * 12.01g = 35453.52g
H (Hydrogen) 4664 * 1.01g = 4710.64g
O (Oxygen) 832 * 15.999g = 13311.168g
N (Nitrogen) 812 * 14.01g = 11376.12g
S (Sulfur) 8 * 32.06g = 256.48g
Fe (Iron) 4 * 55.85g = 223.4g
1 mole of Hemoglobin = 65331.328g of Molar mass
Exercise 126 in Chapter 6
Because of increasing evidence of
