Answers by visitors named: K

thank you so much!!! you really helped!

it has to be choose all that are possiblities. Do you think that a and b then are the best choices? E and G def. are not, agreed.

A&B worked...thank you

My question doesn't have to be answered. I figured out what they were asking and got it correct.

does this sound correct? 8.00X10^-4 kg?

THIS time I got it correct! answer = 5.13x10^-19 molecules! THANK YOU soooo much!! I only had 2 attempts remaining!

it worked out beautifully. Thank you!

I kept trying it, via comparing it to a, & finally did come up with 0.255mol AlCl3...I had just gotten the question incorrect,somehow, 4x...so I had to ask. Thank you for your patience with me. I will be looking into getting a tutor shortly.

I tried this answer & it was incorrect.

ok...so I shouldn't have multiplied through by 2. That makes a lot more sense. Thank you, yet again, and it came out alright.

(oh...slightly dislexic...yes...I meant 0.225 moles, that would have helped if I had noticed before & typed it in correctly.)

this answer, 83.9mL worked...thank you for the guidance.

so for #14 I got: 4mols #16: 1.0mL #17...the other two choices were b-0.1MCH3COOH & then d) both A&B, that's why I didn't bother putting them in, considering b is not an electrolyte. #25: b- HCl? #2, about Av's #...the options were: a)the # of Hydrogen atom in exactly 1g of hydrogen-1 b)the # of carbon atoms in exactly 12g of carbon-12 c)the # of carbon atom in exactly 1g of carbon-12 d)the # of hydrogen atoms in exactly 12g of hydrogen-12 e)the # of moles in exactly 1g of carbon-12. thank you for your time again.

Thank you muchly!! (the questions are just from an old MT that we were to go over to study for the real MT this coming Sat.)

thank you!

#2= 1.61atm:)

ok...I tried my answer...I got the first part correct..it was actually 0.455atm.

for Part 1: would the P2 just be 0.454 atm then? I went: P1*V1=P2*V2 then: P2=P1*V1/V2? So...1atm*1L/2.20L = P2= 0.454atm??

the only prob left is: do I just add the variables? or which ones do I take into account? PV=nRT n=RT/PV...or should I be using: M=mRT/PV???

well...I managed to use the correct equation: M=mRt/PV...& the answer came to Ar. Thank you very much!!

I tired 7.79g of He added & it told me this: This number looks like the final mass of He multiplied by 2. Instead, find the difference between the final and initial masses.

I worked the info above out &, yes, your thinking was indeed correct. Thank you heeps.

take some almonds as a snack throughout the day. As for lunch, take a balanced lunch = with some fruit for dessert, ex.an apple or pear or something of the sort. Main course could be something like a turkey sandwich on whole wheat, or on a whole wheat begal if you so choose. You could also add a cup of cottage cheese in there. They now have those little cups that come like some yogurt does, added with pineapple or strawberry, so that you don't have to worry so much about the flavour of the cottage cheese...if you have a problem handling the 'original' CC. You need the protein to keep you satisfied & to give you enough energy. Sipping on straight water, generally cooler, will help with any undo cravings. According to studies, when you feel as though you are hungry you are more than often just in need of water. Just don't snack on a load of sugar...that won't be of any help.

i mean... 0.574L didn't work.

thank you..I reworked it & the answer came to: 0.73%.

I used: E = -(Rhc)/n^2 R= 1.097*10^7 m^-1 h= 6.6261*10^-34 J s c= 3.00*10^8 m s^-1 E= -(1.097*10^7)(6.6261*10^-34)(3.00*10^8)/16 E= -1.36*10^-19 J so I just use final minus initial? How so? when I do the same as above...but put in 1squared=1, rather than 4squared=16.. I get: -2.18064951*10^-18 then should the answer be: -2.18065*10^-18 - -1.36*10^-19? = -2.04*10^-18??

guess it was correct... the real answer, rounded though, came to be: -2.05*10^-18J thank you.

1. secret 2. search 3. new finder?? 4. dwindce?? dwinced??...not sure if those are words so much.

so...would I find the answer to be: lambda=6.626*10^-34 J s/(0.0459kg)(72.0m/s) answer= 2.00*10^-34 or should I have squared the 72.0m/s velocity?

2.0*10^-34 was correct. Thank you.

the answer came to 3.2*10^-35m. I kept typing the values into the calculator wrong. I multiplied the 765m/s, rather than dividing. I still don't understand the other question though. Q#2.

Soooo??? KE = hc/lambda - 1.602*10^-19J*956eV?? 6.6261*10^-34*3.0*10^8 / 9.7*10^-10m minus 1.53*10^-16?? =5.19*10^-17??? That can' be right??

apparently 5.18*10^-17 kJ/mol isn't the correct answer either.

where am I going wrong?

so just divide by 1000? I tried that when I had 5.19*10^-20 & it told me I was incorrect? or did you mean to divide somewhere else along the way?

WOOO! 3.12*10^4 came to be correct indeed! Thank you ever so much for your time & patience.

terrestrial

Z wasn't correct. Element Z has a larger radius than elements X and Y. Because the outer electrons of element Z are not attracted as tightly to the nucleus as the outer electrons of elements X and Y, we can expect element Z to have a lower first ionization energy than elements X and Y. would it be: element X then: X=105pm?

ignore the "At"...it was supposed to be "Ar-argon"

Place the following elements in order of Decreasing atomic size: tellurium, bromine, argon, cesium, strontium, and selenium. would the correct ranking then be: Cs, Sr, Te, Se, Br, Ar ??? or am I still way off??

the 4th one was correct as well. Thank you!

this order came up being correct. Thank you

I am going to take the above down. Again, thank you.

this ordering came up as being correct. Thank you for your time, it's much appreciated!

you have them as: 0.98, 1.14, 1.17, 1.36, 1.91, 2.35 so...does this mean that: Cs=2.35, Sr=1.91, Te=1.36, Se=1.17, Br=1.14 & Ar=0.98 ??? Therefore this order is then correct, as in being listed in order of Decreasing atomic size?? (this isn't the same as in order of decreasing atomic radius, or is it?..if it is not...I think that's where I messed up before)

Well, I must say then... to all of you whom help each and everyone of the students who use this site, myself included, are very incredible & extraoridinary for taking time out of your busy lives to help everyone!! Thank you to all!!! It does not go unnoticed!!

You are all very welcome, it was very sincere. The part about the Girl Scouts, aside from all of your other work tremendous work, is terrific! (I was a Girl Guide for 10yrs & my mother has been a main GG leader for the past 19yrs now.)

I tried 1s2, 2s2, 2p6, 3s2, 3p6, 3d2 as well, it did not work either.

I ran out of attempts. I am going to attempt to ask under the "help" to see what they say.

I figured out this problem. I ended up just putting the charges on each atom.

"C"...sp^2 was correct, i couldn't find what this compound was either.

I got the answer correct!!...It was BP-BP < LP-BP < LP-LP

I tried for "d" 90-degrees only & it stated the following: Also consider the angle between the equatorial atoms. would this then be: "f" 90 & 180 degrees?? I don't see where 120-degrees plays in this structure & I am obviously missing another to add to the 90degrees, I think.

Ok...I really don't have much of a clue now. "f" was also incorrect. So...does this problem need me to add 120-degrees in there then as well: "e"- 90, 120, & 180...or am I seriously still way off!!!

The answer was "e"...I had to include the 120-degrees as well.

If the four F atoms are in the equatorial positions and the two Cl atoms are axial, all polar bonds cancel. This is what it said when I had tried "always polar"...I had selected the wrong answer. So...it is either then, as I had originally thought: "always non-polar" or "#3...depending on the arrangemen of the outer atoms, this molecule could be polar or non-polar". I am thinking that #3 might be correct now, considering what was written under when I got it wrong. It usually doesn't say anything if you are 100% wrong. Pls help.

yes...I guess they were looking for #3.

is trigonal planar bent though? or would it basically look just as you have it, but with 2electrons on top of the "O" & 2 on the bottom of it? & then 6/3pairs of electrons around the "Cl". or...being as how these structures are way too hard to type out: the "Carbon" would be the central atom with the "Oxygen" sticking straight up above it with one pair of electrons on either side of it (to the right & left) & then the "Hydrogen" atom & the "Chlorine" atom being singly bonded to the "Carbon" & both arranged under the "Carbon" like an upside-down "V"-shape???

I tried drawing as this & it turned out to be correct. Thank you very much!

The answer was B

I will send a msg to the site & see what I receive back. I figure that D should have been correct initially, but I picked C instead by accident...it came up with the feedback stmt & I got more confused.

& for #2 (with the gold/Au): Volume = (407pm)^3??? = 67,419,143 (or: 6.74 * 10^7)?? then does 6.74*10^-7pm's translate to: 6.74*10^+10 cm's?? Then go something like: (4 atoms)(196.967g/mol Au)/6.02*10^23 atoms mol-1 = 1.30875*10^-21 g?? Then density = mass/volume: d = 1.30875*10^-21 g / 6.74*10^+10 cm's *****d = 1.9418*10^-32 g/cm^3????!!!*** PLEASE let me know if I have totally worked these two problems wrong!!! I am pretty sure I am not correct.

3/5 = x/15 find the common denominator which will be "15" then, since you multiply the "5" by "3" to get "15", then you must multiply the top, numerator, by the same. Therefore: 3*3 = "9" If you had: 4/5 to get a full set of 15 counters would be "12"counters...you would have to multiply the "4" by "3" as you did for the above. I think this is what you are looking for?? If not, I appologize.

I would go for capatilizing the "C" on "Club" in this case. If you are saying: "I attended a Christian club," then the "c" would be lower case. However, if you are saying: "I go to Christian Club," then I "believe" that "Club" would be capitalized as being a proper noun? "Christian" in this case would be the adjective...describing the type of club.

Thank you for your time and attention in this matter.

perhaps attempt to show him some practical/real world examples where understanding simple division is a tremendous asset. Go over dealing with $, that may get him interested...or find something that he is a huge fan of & somehow apply division to that. You could also make up work sheets & then time him. If he gets them all correct, or the majority, in x-time, then treat him to a desset or his pick of video to watch, or some play-time. I don't even know what kids in the 4th grade do anymore, but find some sort of slight "star" to congratulate him for his hard work & efforts.

this worked out perfectly!...thank you...that was the last question for the year that I had to finish through that program. I really must not have been paying attention with "a*2", as opposed to dividing a/2 to solve for "r". Thank you again, & for having pointed that out & with the conversion of pm's to cm's. That looks much faster & correct! Thank you for all of your help, & bobpursley's help this past semester!!! It all has guided me to understand what I am doing, for the most part, sooo tremendously!!!!!

oh yes...the answer should look like this: r = _____pm

I think that I used Robidium's atomic mass instead of Rdodium's...I'll check now if that works better. Does it look as though I am on the right track though??

I tried exactly what I had done with the above & got my answer to be: 190.2pm's & was still wrong!! Could you please guide me as to what I should be doing? I only have 2more attempts left at the question.

always should be an adverb...describing what the action is which is the verb "late". Try that, it "should" be correct.

I ran out of attempts but I still would like some help trying to figure out how to do this question properly, please.

I did manage to get as far as the 380.5pm's. I guess where I did go wrong is that I missed carrying it further than that. I missed the 380.5*sqrt 2=538.1pm's & then the remainder of the answer. Thank you very much. It's good to know that I was getting the handle of it, aside from the last part, at least.

how about these two Crocodiles,quiet as logs,lurked on the riverbanks. Large and small dinosaure,stalked the grassy plains.

i still didn't get it is the subject you or the shoe shop. even the other one i didn't get it

should it be: 2-chloropentane & 1,2-dichloropentane??? ie. "D" ??? I don't think that it could be "I or III"

I think that it must be "a", too, when I rethought it. Thank you.

ignore this post.

ignore this post.

(note: cyclohexanol is a hexagon with an "-OH" attached off one of the carbons.)

cyclohexanol: OHCHCH_2CH_2CH_2CH_2CH_2 (I think that is sort of it, otherwise.) Or just: C_6H_11-OH

The experiment is on dehydration of an alkene & then to test it to see if it is "unsaturated" or "saturated", which I really should have specified!!...sry!!!! For this exp. we are to only have like 3 drops of the Br in CH2Cl2 & add in 1-2 drops of cyclohexene (the product from Cyclohexanol)...then see what happens & attempt to write an equation...which is really that part that I am not sure on whatsoever!

Thank you both of you for your help.

answer for the first was "3"

"A" was correct

Making the decision to return to school was not an easy decision; however, returning to school will create a better future for my family and I. Although seeking to attain a college degree while having to work full-time and raise children, will be a defininte challenge, this challenge will be worth it (try to find a better word than "it" though...if that doesn't work for you so much) to ensure a brighter future for all of us. Learning new skills and knowledge is required everyday in my job. At present, I am not eligible for promotions due to not holding the necessary college degree. Also, when I try to compete with others for jobs that do not require a college degree, the other applicants have some type of formal education on their résumés that outweighs my background. I desire to attain a college degree so that I am able compete with others for the positions for which I am qualified. Today’s employment market is filled with talented people who have various degrees, speak multiple languages, and have varying experience levels. Without a degree, it will become more difficult as I grow older to find a position I not only enjoy but one that also provides for my family. 1. MY EDUCATIONAL GOALS I am not sure if this is to be an actual "formal" letter for a very specific reason...but, I have rearranged just a few areas. Check to see if those help somewhat. Perhaps someone else will also be able to add/edit your piece further. If this letter is being written in hopes of being selected for post-secondary Ed, of some sort, perhaps throw in the words that show desire & dedication or persistence. If the above is a formal letter, try not to bring what others have over you quite as much, especially if it for applying to further you education. Just sell yourself. Somewhere you "may" have wanted to state about furthering your education in hopes of....etc...rather than just calling school "school".

take out "Making" at the beginning, also. Just begin with "The decision". Take out the comma I put in after children & before will in the 2nd sent.

( "B" is meant to read as a cyclohexane with a double bond off of one carbon & then a CH2 at the end of that)

<cyclohexane>=CH2

molecular compounds

Thank you...good to know, but isn't there a "Dr.Russ" that works with Organic Chem??

thank you

so then "B" as the "Major" product? -the cyclohexane with the double bond off from it & then the "CH2" attached at the end? What is the correct name for this compound anyways?!

just saw it - thank you.

Thank you for all of your information! So then I believe that the answer "should" be: "C" (3-methyl-1-cyclohexene) b/c then the double bond would link to where the Br was originally & flip backwards..."a" (1-methyl-1-cyclohexene) could be a "minor" product then (correct?) & I don't think that it could be "d" (4-methyl-cyclohexene). what is the compound called with the cyclohexane with a double bond CH2 attached at the end called exactly?...I have had other questions showing it & I don't know the exact name of it.

So do these look to be as though I am on the right track with my answers to the above questions??? Except, for the FIRST question...I think that #3 should be SN2, rather than SN1 (substitution reactions).

or even just: Is it still okay to conclude a formal essay using something like: "Thus, in conclusion, ..."; or, are there much better/higher levels to go about writing a proper conclusion...or even just a basic thesis stmt?

got it. Thank you for you help. I don't know what i was thinking when I wrote down my answers to which is the best leaving group, but least to easiest. I meant to write them both a beginning with F, Cl, Br, then I. Back to that other question, re: MAJOR product formation. So...then the idea that I had about "A" occuring only in MINOR amounts is reverse then & it would then be the MAJOR product?...ie...so then it would be 1-methyl-1-cyclohexene, as opposed to 3-methyl-1-cyclohexene (this one would be the MINOR then???). I have been quite confused when it comes to the MAJOR/MINOR product formation of chemical reactions.

it is balanced

for the reactants you have N=1 & O=4. for the products you have N=1 & O=4.