Questions and Answers
Your Questions About X100
What to expect when boarding at a private school?
I am planning on boarding at a private school. I like the idea of sleeping at school because I wont have to worry about being late. I would like to know what to expect in terms of freedom, getting to see my family, waking up for school, and the advantages of being a boarding student.
not a lot. There are lots of rules about curfew, over-night permissions, etc. They watch over you like a hawk so they know you are safe and sound. But most schools allow you to do whatever you want in your freetime, including going into the nearby town.
Getting to see your family:
it depends on how far you live but most kids see their family every month through holidays, long weekend, parents weekend, college info weekend, etc. Internationals go home about 3 times a year
waking up for school: i wake up an hour before class but i know some boys roll out of bed 10 minutes before class and still make it. (but smell bad)
advantages of being a boarding student:
the experience of having a roommate, dormmates, making the closest friends you have ever had. For boys it means x50 the number of video games, etc and for girls it means x100 more dresses, shoes, jewelry you can borrow haha
you also dont have to spend time commuting to and from home so you can sleep instead
expect to become a procrastinator and an awesome BS-er
What is the best camera for a beginning photographer?
I am looking for a camera that is light weight, smallish, and durable but hi-res and clear. I am just getting in to photography, and i definitley want ditigal.
There are lots of good brands out there, but the best feature to look out for is the “Manual” mode (this lets you set aperture and shutter speed manually). This feature would help you a lot in learning the basics of photography.
I’m personally a Canon fan, but for a compact digital camera I’d go for a Fujifilm FinePix X100. You can also compare cameras side-by-side on the DP Review website
What percent of the population are heterozygous if the population is in Hardy-Weinberg equilibrium?
In a population of flowers, the allele for purple petals, P, is dominant to the allele for white flowers, p. If 84% of the population has purple flowers, what percent of the population are heterozygous if the population is in Hardy-Weinberg equilibrium?
Okay firstly I’m renaming the allele for purple flowers to W which is dominant to the allele for white flowers w. I’ll do it slowly so you can do your next problem yourself!
So to be white you need the genotype: ww because it’s recessive.
To be purple you can EITHER have the genotype: WW or the genotype Ww because it’s dominant.
It tells us that 84% of the population is purple. In other words; “84 ‘out of’ 100 individuals are purple” OR “in a population of 100 individuals there will be 84 purple flowers”. So the frequency of purple individuals in this population is 84/100 which simplifies to 0.84/1 which = 0.84.
So we know that 0.84 have either the genotype WW or Ww. In the hardy-weinberg equation we know that p^2+2pq+q^2=1 in other words; the frequency of homozygous dominant individuals plus the frequency of heterozygous individuals plus the frequency of homozygous recessive individuals is equal to 1. We know that all the individuals which are p^2 are purple and all the 2pq individuals are also purple. So if we add these we will account for all the purple individuals in the population. So; q^2+2pq = the frequency of purple individuals in the population which equals 0.84. Now we replace q^2+2pq with 0.84. When we re-write the HWE equation, we get; 0.84+q^2= 1 we can re-arrange this to; q^2= 1 – 0.84 which is equal to 0.16. So q^2 = 0.16. Now we can find the allele frequency of the w allele by taking the square root of q^2. So the square root of 0.16 is equal to 0.4.
We also know that p+q=1. And we know that q is equal to 0.4 so p+0.4=1 we can rearrange this to; p=1-0.4 which is equal to 0.6. So the allele frequency of p is 0.6. Looking back at the HWE equation we know that the frequency of heterozygotes is equal to 2pq. So if q=0.4 and p=0.6 then 2×0.4×0.6= the frequency of heterozygotes which is equal to 0.48.
So 0.48/1 = frequency of heterozygotes. In percentage form we want the answer ‘out of’ 100. Currently it’s ‘out of’ 1. So to get from 1 to 100 we multiply by 100. So (0.48/1)x100 =48/100 = 48%.
We can check our answer by using the HWE equation; p^2+2pq+q^2=1 which is 0.6^2+0.48+0.4^2 <—-this should be equal to 1 if we have done everything properly. So; 0.36+0.48+0.16 which DOES equal 1. Take a pat on the back and a sigh of relief.
What percentage of the Naphthalene molecules emitted a photon?
A 8.0 mL ampule of a 0.110 M solution of Naphthalene in hexane is excited with a flash of light. The naphthalene emits 11.1 J of energy at an average wavelength of 349 nm. Best answer will be given to who can show how to do it as well as give the correct answer!
First use the formula E = hc/?
6.626*10^-34 x (times) 2.998*10^8/ 349*10^-9 nm = 5.692*10^-19J
# of the photons emitted
11.1 /5.692*10^-19 = 1.95*10^19 Photons
moles of nepthalene
8/1000 x 0.110 =8.8*10^-4 mol
# of its molecules
8.8*10^-4x 6.022*10^23(avogadro #) =5.30*10^20 molecules
make sure double check the calculation
Powered by Yahoo! Answers