dg2008

by converting 1x10^3 kg m^-3 to km. see attachment. then 1x10^3 kg / 1×10^-9 km^3

volume x density so I have 5.0x10^-4 km^3 as the volume and 1x10^12 kg km^-3 as density but I'm confused about the measurements for each. Does it become kg km?

Can someone please explain to me the process of cracking oil to produce ethene or propene.

Question Amongst the radioactive elements that were incorporated when the earth was formed were: uranium 238U with a half-life of 4.5 × 10^9 y uranium 235U with a half-life of 7.1 × 10^8y thorium 230Th with a half-life of 8.0 × 10^4 y iron 60Fe with a half-life of 1.5 × 10^6 y manganese 53Mn with a half-life of 3.6 × 10^6 y thorium 232Th with a half-life of 13.9 × 10^9 y or 1.39 x 10^10 Given that the earth is believed to have formed 4600 Ma ago, which of the above could still be contributing to the radioactive heating of it's interior? Answer In that case 4600 Ma = 4.6 x 10^9 y Start by calculating the ratio of what is left, that ratio for what has decayed is 0.5 raised to the power time/half-life. The time being the age of the Earth, so it's the reciprocal of that. So for the first one that's 0.5 to the power (4.6x10^9/4.5 x10^9) or 0.49 take the reciprocal and 51% of the original U238 remains. Am I going in the right direction? I'm thinking that anything below 10^9 can be ignored.

Question Explain the surface process by which material that originated in grainte may eventually accumulate as a mud. Weathering, erosion, transportation and deposition (750 words) So far I have: Granite, an igneous rock is formed deep within the Earth and cooled very slowly, to contain large crystals; quartz, white feldspar and mica. Physical weathering, heating & cooling produce forces in the rock, expansion & contraction stress the rock. Eventually the forces begin to split the mineral form each other allowing water to seep along the grain boundaries. Freezing and thawing then liberates the grains. Once it's broken up, chemical weathering gets to grips on the surfaces. The main processes of chemical weathering are acids and oxidation, which often operate together. As rainwater dissolves carbon dioxide it is slightly acidic, this attacks the minerals in the rock. The feldspars and the micas are very susceptible to chemical weathering. The quartz is very resistant to chemical weathering. During weathering, hydrogen ions break free their loosely bonded metal atoms. When this happens the structure collapses to form clay minerals. The minerals are exposed to flowing water that lead to erosion. The grains as they are carried away pound, blast and scrape more debris to wear away the surface when they pass. The grains are also broken down as they move along to become smaller and more rounded. Transportation of grains or sediment depends on the energy supplied by the transporting system and the size of the grains. Large grains settle to the stream bed, smaller grains roll or bounce along the stream bed and even smaller grains are picked up and carried along in suspension. The grains are also exposed to physical abrasion and chemical attack. Reactive minerals form clay minerals and dissolved ions while quartz is liberated. The transported grains eventually collide with the stream bed and each other. The rock fragments tend to become rounded by the abrasion, rather than broken by impact due to the water cushioning the impacts. ----- Thats about it. For deposition should I talk about currents, flow rates etc, degree if sorting? Also should I add more detail to above? like chemical equations, examples? I think my main problem is refering it to mud.

so: n = 0 charge p = +e charge n + p → p + p + π– = 0 + e → + e + e + -e = 1e n + p → n + p + π+ + π– = 0 + e → 0 + e + e + -e = 0 n + p → n + n + π0 = 0 + e → 0 + 0 + 0 = 0 n + p → n + n + π+ + π– = 0 + e → 0 + 0 + e + -e = 0 so the first one is the only one that could work. 3rd due to energy, 4th due to both energy & charge, or am i completely wrong

Question A neutron and a proton collide with a kinetic energy of 150 MeV. You may assume that a neutron and a proton each have the same mass energy. Given that a pion has a mass energy of 140 MeV, which of the following reactions are allowed and which are not allowed? n + p → p + p + π– n + p → n + p + π+ + π– n + p → n + n + π0 n + p → n + n + π+ + π– Reaction allowed? If not is it due to Energy conservation &/or Charge conservation I had thought that the middle two were allowed How do I work out the conversions?

Start by reading Experiment 1 and then use your knowledge and understanding of the processes of photosynthesis and respiration and the properties of enzymes as biological catalysts to write the interpretation of the results for Experiments 2, 3 and 4 which follow. For each experiment your answer should be less than 200 words. Experiment 1 Method A fixed quantity of Elodea canadensis is placed underneath an inverted water filled funnel at 20 °C in a beaker of pond water. A bright light is shone on the beaker. After a few hours a collection of gas is apparent in the test tube as the water it contained is displaced. The experiment is left to run for a specific time (6 hours) and the volume of gas is recorded. pond water gas collecting support to keep funnel off bottom bright lamp Elodea canadensis Results 10 cm3 of gas have collected in the test tube; the gas is identified by simple tests. Interpretation In Experiment 1, the Elodea canadensis is photosynthesising. Light and carbon dioxide (dissolved in the pond water), and water are available to the plant and it is able to synthesise glucose by the process of photosynthesis. The gas produced is oxygen. The plant will also be respiring: that is reacting glucose with oxygen. Photosynthesis must be occurring at a greater rate than respiration, as there is net production of oxygen. Experiment 2 Method Experiment 1 is repeated but this time in the dark. The apparatus is left for the same amount of time as Experiment 1 and the volume of the gas is recorded. Results 2 cm3 of gas have collected in the test tube, the gas is identified by simple tests and is found to be different from the gas obtained in Experiment 1. Experiment 3 Method Experiment 1 is repeated at a range of different temperatures. For each temperature the amount of gas produced over a fixed time period is measured and recorded. The data collected is used to draw a graph of the rate of gas production against temperature. Results Graph 1 A graph of rate of gas production (measured in cm3 h−1) against temperature for Elodea canadensis. Experiment 4 Experiment 1 is repeated once again but this time carbon dioxide is bubbled into the water so the amount of dissolved carbon dioxide in the water is increased above the level used in Experiment 1. Results 20 cm3 of gas have collected in the test tube; the gas is identified by simple tests to be the same as the gas produced in Experiment 1. ----- Experiemnet 2 is about respiration or dark reaction & carbon dioxide fixation? Experiment 3 is talking about rate of reaction with increasing temp experiemnt 4 is about in increase in Oxygen due to increase in Carbon Dioxide.

here is the other information I am given: Figure 2 was produced by Peter Grant, the expert on Darwin’s finches, to illustrate his book on the ecology and evolution of the finches. He and others measured the beak (bill) depth of a large number of specimens of three species of ground finch from three different islands. Beak depth is grouped into categories of 0.25 mm, so that, for example, all birds with beak depths between 3 and 3.25 mm fall into the same category. The y-axis of the figure shows the frequency with which each beak depth category occurs in the population. For example on the island of Los Hermanos, the frequency of a beak depth of 3.5 to 3.75 mm is 0.42. This means that 42% of the sample measured had a beak depth within that category. If the total number of birds examined at each location is large enough it will give a mean that is close to the mean that would be obtained if every single finch in the population was measured. Since the sample in this case was a large one, it is likely to be representative of the whole population. The mean beak depth for each species on each island is indicated by an arrowed vertical line. and having just read the last bit again I take it I just need to read of the line

(a) I'm confused CC & CCr? or am I looking for the actual Palomino so CCr & CCr (d) CC x CCr = CC CCr CC CCr CrCr x CCr = CCr CrCr CCr CrCr CC x CC = all CC CCr x CCr = CC CCr CCr CrCr CC x CrCr = all CCr CrCr x CrCr = all CrCr

can you give me an example? I'm feeling so thick today lol so 2.5 x 0.08 (first bar) + 2nd + 3rd etc = total / total frequency (0.98)

by 'divide by number of values' you mean number of bars?

how do I work out the mean from these figures?

Palomino horses are valued for their beautiful and distinctive coat color; usually a golden yellow or shade of tan with a white or flaxen mane and tail. They result from matings between chestnut coated horses and ‘off white’ coated horses and the palomino coloration is actually intermediate between the coat colors of the parents. In a breeding experiment, several pairs of palomino horses were bred together and produced 19 ‘off white’ offspring, 21 chestnut offspring and 44 palominos. (a) Determine the genotype of the palomino parents used in this breeding experiment and explain how the presence of these alleles in the genotype contributes to the phenotype. (b) Show the possible genotypes and phenotypes with respect to coat colour of the offspring produced by matings between palomino horses. © Calculate the expected ratio of each of the phenotypes in the offspring and compare this to the actual ratio of the different phenotypes produced. (d) A horse breeder wishes to establish and maintain over many generations a herd with as many palomino horses and as few ‘off white’ horses as possible. She wishes to breed from her own horses only. Explain which phenotypes of horse from the herd should be used for breeding in order to maximize the number of palomino horses and minimize the number of ‘off white’ horses in the herd over many generations. ------------- my attempt (a) CC & CrCr (b) chestnut = CC, palomino = CCr x2, off white = CrCr © haven't done yet (d) breeding CCr with CCr produces CC CCr CCr CrCr so 2 Palomino and 1 off white. Breeding CrCr with CC produces CCr CCr CCr CCr so all Palomino

Does anyone know the genotype/phenotype of the following: off white, chestnut & Palomino horses - all bred from Palomino horses) chestnut = ee, aa, CC, dd, gg, ww, toto palomino = ee, CCr, dd, gg, ww, toto off white = ee, CrCr (Chart based on Dr. Bowling's Genetic Formulas)