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hey. see the attachment
I am just going to post the answers if you have questions then ask...
1) I believe it would be thrombocytes cuz the innactive form is platelets...
2) 4 (2x2 heterozygotes =4)
3)I didnt get any of the answers you provided, I got (.3)(.05)(44). because you have .05 mol/1L multiply by .3L you get moles and then multiply by 44g/1mole and you will get just grams...the question is asking for grams so if you do this the liters will first cancel and then the moles will then cancel when you multiply by the MW...
4) I got 43.5g of NaCl...dont understand how i got this one wrong cuz its not even an answer choice but i am pretty certain its correct...i used vant-hoff factors for this...2 for NaCl and 3 for SiO2
5) im gonna guess a basic solution of 9...since the molarity of NaOH is higher than that of 90% .1M HCl which is .09M...
6) 2nd order i believe
7) 2???
8) sp3
9) B, (60)(12)/(.1)
10) ill go with A...OH- is a greater electron donating group than CH3 and benzene...higher the electron donating effect (stronger activating group) the more easily it will react with an element (Br-) that is looking for electrophillic substitution/addition to the ring...
Please correct me if iam wrong on number 10 thanks!!!
I am just going to post the answers if you have questions then ask...
1) I believe it would be thrombocytes cuz the innactive form is platelets...
2) 4 (2x2 heterozygotes =4)
3)I didnt get any of the answers you provided, I got (.3)(.05)(44). because you have .05 mol/1L multiply by .3L you get moles and then multiply by 44g/1mole and you will get just grams...the question is asking for grams so if you do this the liters will first cancel and then the moles will then cancel when you multiply by the MW...
4) I got 43.5g of NaCl...dont understand how i got this one wrong cuz its not even an answer choice but i am pretty certain its correct...i used vant-hoff factors for this...2 for NaCl and 3 for SiO2
5) im gonna guess a basic solution of 9...since the molarity of NaOH is higher than that of 90% .1M HCl which is .09M...
6) 2nd order i believe
7) 2???
8) sp3
9) B, (60)(12)/(.1) i thin the answer is E. because you want to find out how much you need to add to 60mL so we have to subtract V2 from 60.
10) ill go with A...OH- is a greater electron donating group than CH3 and benzene...higher the electron donating effect (stronger activating group) the more easily it will react with an element (Br-) that is looking for electrophillic substitution/addition to the ring...
Please correct me if iam wrong on number 10 thanks!!!
AGAIN can some1 please tell me when u have to subtract your V2 from the initial volume....i can not get this straight in all of achiever and top score u never did that i was looking for both types of question so i could compare but didnt happen....what is the "word" or "phrase" your looking for that you subtract the initial from the final....and what is the "word" or "phrase" u look for when u just leave it.....can some1 give examples of both PLEASE dat in 23HOURS!
5) im gonna guess a basic solution of 9...since the molarity of NaOH is higher than that of 90% .1M HCl which is .09M...
The answer is pH=7, he uses the word neutralized
If ""90%"" of a 0.1 M solution of HCl is neutralized with 0.1 M NaOH, what is the pH of the solution?