Answers by visitors named: ~christina~

Is there a question? It seems as if this is just a formula.

Alright well after thinking about it you may have wanted R from the formula. If this is so then this is how you'd find it. divide the PT over to the left you can do this with both P and T since they are being multiplied so you'd get I/PT= R =)

Alright then =)

It's because of the absolute brackets. The absolute brackets | | mean that it is either 20 or -20 so using the -20 and solving for z would give you the book's answer of 3 but using the 20 would give you -2 for z

D=M/V Just my calculations.. empty flask + stopper= 37.34g flask + stopper + water= 63.67g water= 63.67g - 37.34g= 26.33g D of water = 0.9970g/cm^3 V of water = M/D V water= 26.40cm^3 flask + stopper + liquid = 52.01g liquid= 52.01g- 37.34g= 14.67g V same as water so...26.40cm^3 then to get D for unknown liquid D=M/V D= 14.67g/26.40cm^3 = 0.5556g/cm^3 Unknown Metal D (have to find the volume) stoppered flask + metal = 61.28g stoppered flask empty (from before)= 37.34g metal mass = 61.28g- 37.34g = 23.94g Stoppered flask + water= 78.19g water mass = (78.19g + 23.94g)-37.34g= 67.79g Vol of water = Mass/Density 67.79g/(0.9970g/cm^3)= 67.99cm^3 Orig Vol- new Vol= 67.99cm^3-26.40cm^3= 41.59 cm^3 (vol of metal) D of metal= Mass/Vol 23.94g/41.59cm^3= _________g/cm^3? Hint: it's not the same =D

The correction for this(If anybody cares to look anymore) (I feel I have a responsibility to correct the error =() So... _______________________________________ water mass = 78.19g - 23.94g-37.34g= 16.91g Vol of water = Mass/Density 16.91g/(0.9970g/cm^3)= 16.96cm^3 Orig Vol- new Vol= 26.40cm^3 -16.96cm^3= 9.44cm^3 (vol of metal) D of metal= Mass/Vol 23.94g/9.44cm^3= 2.54g/cm^3? I checked twice this time so it SHOULD be correct.

Yes Now looking at that again... it should be subtraction. Thanks for catching that Dr.Bob =D (I should check twice next time)

10,000 x .10 = 1000 interest a year 5 years x 1000= $5000 10,000 + 5000 = $15,000 (If you didn't spend any of the 10,000 of course lol.

I don't like to butt in but I have done this lab during thes ummer and the steps are in my lab manual. Preparation and Standardization of HCL with Sodium carbonate (Na2CO3) (in general without being specific as to the ammount or anything) ( I actually had to go and mix concentrated HCL not knowing the Molarity but I'm ommiting this step) - weigh out Na2CO3 into 3 flasks, and dissolve in distilled water - add 3 drops methyl orange indicator into solution - titrate with HCL solution untill solution just changes from yellow to red - boil sol for 2-3 min, cool to room temp and continue titration to endpoint =)

No problem, I help where I can =D

Well...I corrected what I did wrong before and from that ... to see what's happening with your calculations.. (23.94g)/(0.9970g/cm^3)= 24.01cm^3 then using this answer I need to caculate the Density of the Metal: (23.94g)/(24.01cm^3)= 0.9970 g/cm^3 ______________________________________ The density is NOT the same as water. so it would not be the mass of metal divided by density of water. It would be however, what I later posted on the last last post. Orig Vol- new Vol= 26.40cm^3 -16.96cm^3= 9.44cm^3 (vol of metal) D of metal= Mass/Vol 23.94g/9.44cm^3= ____g/cm^3

Your very welcome, I actually had a hard time figuring out that a few years back as well so I know what your speaking about =D

For the first part: births add to the population deaths subtract from the population migration can be a subtraction or addition to a particular population

factors affecting these rates could be natural disasters which could in turn affect the ammount of food and thus affect the death rate/migration rate/ and birt rate

A suggestion is to go and google how to find a common denominator in fractions click on the first site on the list and it should help you with the lowest common denominator hope it helps =D

I meant the 2nd site

Thanks for commenting on you own experience in hs. I didn't actually have AP chem in my school that I know of. The only AP classes in my hs was History, Calculus, English, Bio and the year after I graduated they added Physics. I took AP history and English but didn't score high enough for college credit. That test is expensive I must admit though. What kind of labs did you do in hs? I bet it wasn't as complicated as in college right Dr.Bob?

Well Bob we didn't need to actually do experiments on that test so I guess lab wouldn't have been necessary. However in bio we didn't need to go and dissect a frog to know anything for the test but we did so anyways. Not a fan of AP classes? Well the setting of the normal classroom in highschool and having constant quizzes really doesn't imitate the college experience at all I must admit. Especially since in college we only have 3 tests each semester in general.

Basically we learned through books and pictures and just did practice with problems but not through actual aplications in lab. (we talked about things that happen in experiments like we mix stuff together and what should theoretically come out of that without actually doing the experiment)

If your speaking of AP classes hosted by a college then I don't know about that, but I was speaking of AP classes in actually taught by highschool teachers. To get into those classes we didn't have to take a test but rather ask permission from the head of the department (who taught the class) and they also had to review our grades.For history it was the previous grade on the Regents exam. In both my classes we had about 15-20 "select" students. (I seriously don't know how some people got into this class) What I can say about chemistry in highschool on the topic of learning the basics is that I didn't remember anything chemistry related (the only thing I remembered about pH was the word and that it had to do something with acidity) by my 4th year in highschool. I took a year total in chemistry my second yr of highschool and I forgot everything. I personally have to say that you would have to be interested in a subject before you remember anything for a long period of time. If your speaking of basics in terms of math skills, critical thinking and such well I would have to say that I know that graduate school is a while away but I'm kind of aprehensive when I think about the GRE exam since from what I've seen online it's just like the SAT's. I didn't do well on the SAT's maybe b/c I didn't practice but I've never been too good at tricky math problems or critical reading. I think it's WRONG that they go and determine how good a student is by a single exam on things you don't even learn in highschool. My friend did great on the test and I did better than her in class. (I get mad just thinking about it). I don't know if the GRE is very important in entering a grad program but would know by any chance?(even though they probably didn't have this test back then) With my major now I can actually enter medical school (that test by the way is the MCAT) but I don't want to. Wow you must have been really close to your students to know that all of them were admitted to medical school. (I would have to say it's rare as I don't know my teachers and I'm probably just a name on their list) I guess I wouldn't want a carpenter to build my house if all they knew was from observing rather than actually doing. P.S I really don't mind about hearing about what's wrong with chem education but maybe that's b/c I don't have much to do until school starts.lol But if you don't want to elaborate that's fine.

I missed out on something then Writeteacher.

Yep um I said that the test was the MCAT in the post response to the previous post by you... Aww come on..you can't blame tests for the metal detectors.. That is not true Dr.Bob. My point is that people aren't going to bring unacceptable items into school because they are stresed about tests but rather because of the drama of school and their aquaintances. Usually the troublemakers don't exactly excell in school. (I don't want to make that generality but it's true [I used to be picked on by them so I know])

Depends on what the question is and what your graphing.

Let me help you on this... To find the limiting reactant first of all. -You have to first find the moles you have of each reactant. - then go and find out how much of each reactant do you need to consume the other reactant - then compare it to how much moles you do have of each and the one that needs more to consume than you have is the limiting reactant I'll start with this and PLEASE post your work as I'm NOT going to do all the work =) (I'll check later)

For the binomial theorem, typing binomial theorem into Wikipedia should help from what I looked at on there.

to convert to kg 1kg=1000g

Hm..the decimal point? lets see... 12,321g (1kg/1000g)= 12.321kg

Yes this is correct.

hundreds / tens/ ones . tenths/ hundredths / thousandths ex 123.456 1- hundreds 2- tens 3- ones decimal point 4- tenths 5- hundredths 6- thousandths hope it helps =D

well for this question the whole goal is to get all the x's on one side and all the numbers on the other side. 1st combine all like terms on the same sides 5X-9-19+22X-38X=-17+17X-9+12 -11x-28 = 17x-14 then go and bring the x's over to one side and the numbers over to the other side(subtract or add) -28x= 14 then divide to get x x= 14/(-28) so what do you get?

The square root and the 7^2 cancel each other out so all you get is 7 + 3=10 same for every other number as well ex sqrt(2^2)= 2 hope this helps=)

(2x3)^7= 279936

simple form?? just add and subtract, I'm not sure what you mean I get -216 from doing the problem though.

NO NO NO read what I posted below please

Let me help you on this... To find the limiting reactant first of all. -You have to first find the moles you have of each reactant. - then go and find out how much of each reactant do you need to consume the other reactant - then compare it to how much moles you do have of each and the one that needs more to consume than you have is the limiting reactant - use the limiting reactant to go and solve for how much is produced. - convert moles to grams produced I'll start with this and PLEASE post your work as I'm NOT going to do all the work =)

I just gave you instructions for finding the theoretical yield Josh.

-You have to first find the moles you have of each reactant. - then go and find out how much of each reactant do you need to consume the other reactant - then compare it to how much moles you do have of each and the one that needs more to consume than you have is the limiting reactant theoretical yield is: - use the limiting reactant to go and solve for how much is produced. (theoretical yield) - convert moles to grams produced Post work based on steps and if you have problems with the steps just post what you get up to and I'll analyze. =)

I'm not looking at the problem at the moment. Did they ask how much of the NH3 remained unreacted. If so, you will need to calculate how much of the NH3 was used and subtract that from the 30 g with which you started. g NH3 used = 0.0626 g CuSO4 x (4 mols NH3/1 mol CuSO4) = 0.250 g NH3 (actually you did this when you determined the limiting reagent). So we started with 30.0 g NH3. We used 0.250 g and we have left 30.0 - 0.250 = ?? What?? Dr.Bob g NH3 used = 0.0626 g CuSO4 x (4 mols NH3/1 mol CuSO4) = 0.250 g NH3 You used g of CuSO4 to get moles?? But 0.0626 is the moles of CuSO4 found not the grams. Then from what I did before don't you have to go and convert using the ratio to find the moles used up by CuSO4, THEN go and use the molar mass of NH3 to convert the moles of NH3 to grams of NH3 used up ?? HOW did you get 0.250g ...I think I got: ______________________________________ for the grams of excess reagent since you found that the limiting is CuSO4 then what you have left is only NH3 right? to find the amount left: you have 30g of NH3 in the begining so how much did CuSO4 use up? use the same equation to find out the limiting to find out how much was used up (you already found this) 0.2504mol NH3 used up (17.034g/1mol NH3)= 4.26g used subtract the amount you have from the ammount used 30.0g NH3 have - 4.26g NH3 used = ____? _______________________________________ Please check my thinking Dr.Bob =)

If your speaking of moles in the 1st step No you don't have 1mole... starting with grams you find moles: 10gCuSO4 (1mol/159.62g)= 30gNH3 (1mol/17.034g)=

yes but watch the significant figures CuSO4 moles= 0.0626 mol NH3 moles= 1.76 mol

not necessarily you have to go to the next step and find out how much do you need of each to consume the other. use the moles that you found (minding the significant figures) using the equation to see how much the ratio is between the NH3 and CuSO4. #moles CuSO4(#mol NH3/ #mol CuSO4)= moles NH3 needed to consume all CuSO4 you have #moles NH3(#mol CuSO4/ #mol NH3)= moles CuSO4 needed to consume all NH3 you have then compare that to the amount of each that you have and then you see which you do not have enough to consume the other of.

to clarify that the moles of each multiplying CuSO4 + 4 NH3 ----> Cu(NH3)4SO4 #moles CuSO4(4mol NH3/ 1mol CuSO4)= moles NH3 needed to consume all CuSO4 you have (the moles # is the amomunt you found before I just wanted to clarify the moles in the ratio is from the equation and it's the same except you flip the ratio of course to find the moles of CuSO4 needed to consume all the NH3 you have)

Yes exactly Josh.. now using the limiting amount of CuSO4 finding the ammount of product you can produce will be the theoretical yield. so... 0.0626mol CuSO4 (1mol Cu(NH3)4SO4/1mol CuSO4)= ? then taking that ... mol Cu(NH3)4SO4 produced (1mol / molecular weight of Cu(NH3)4SO4) molec weight = add up all the elements involved in the product

Yes... so compared to what you have 0.0626mol of CuSO4 1.76 mol of NH3 If you want to consume all the CuSO4 you'd need 0.250mol NH3 while if you want to consume all the NH3 you'd need 0.440 mol of CuSO4 which one would you have enough to consume and which one do you not have enough to consume the other? (limiting reagent?)

molecular weight is a little off... it's actually from what I calculated 227.756g so the theoretical yield is 0.0626 mol Cu(NH3)4SO4 ( 227.756g /1mol Cu(NH3)4SO4)= ? I'm just going to say I got 14.26g of Cu(NH3)4SO4

Josh, I have to seriously get to sleep so I'll just post this to help you out with the rest and I'll check tommorow morning to see what you post if anything =D for the grams of excess reagent since you found that the limiting is CuSO4 then what you have left is only NH3 right? to find the amount left: you have 30g of NH3 in the begining so how much did CuSO4 use up? use the same equation to find out the limiting to find out how much was used up (you already found this) 0.2504mol NH3 used up (17.034g/1mol NH3)= 4.26g used subtract the amount you have from the ammount used 30.0g NH3 have - 4.26g NH3 used = ____? _______________________________________ for the percent yield: the equation for that is percent yield = actual yield (given)/ theoretical yield Hope this helps =)

Yep I just went back to look since I forgot to say x100 for the percent yield but you figured it out yourself Yes it's 88.4% in significant figures Your welcome, I help when I can of course =)

go and find the lowest common denominator between the 4, 5 and 2 a common multiple is 20 so.. 3x/4 - 2x/5 +x/2= 5(3x/20)- 4(2x/20) + 10(x/20)= then multiply ... 15x/20 - 8x/20 + 10x/20=__________ ??? now add and what do you get?

72a^2*b^3 ___________ -24a^2*b^5 simplifying you go and reduce then divide the larger by the smaller 4a^2*b^3 ___________ -1a^2*b^5 now I divide...a's cancel out and b's subtract since you divide the b's and subtract the larger by smaller b^5-b^3=b^2 -4 ___ b^2

I meant -3/b^2 of course.

separate and multiply based on FOIL first outer inner last (2y-3)^2= (2y-3)(2y-3) then use foil and multiply the first term, outer term, inner terms then last terms . (2y-3)(2y-3) first terms multiplied 2y*2y= 4y^2 outer terms multiplied 2y*(-3)=-6y inner terms multiplied (-2y)*3=-6y last terms multiplied (-3)*(-3)=9 Now combined FOIL (first outer inner and last terms ) 4y^2-6y-6y+9= but now you can combine the -6y's so (-6y-6y=-12y) and you get 4y^2-12y+9

Your welcome =)

inner terms multiplied 2y*(-3)=-6y it is the same as the outer terms multiplied (the negative was misplaced before)

Oh..it was a duplicate? Oh well I solved it Dr.Bob.=D anyways..ha

x-3(2x+1)=? since y=2x+1

x-3(2x+1)=?? you have to solve it by multiplying the 3 over and then just adding the x's and the #'s without the x's Nobody's going to give you all the answers.

well x-3(2x+1)= multiply the -3 and the 2x+1 then add

deoxyribonucleic acid it stores genetic information

well move the x's over to one side and the numbers without the x's over to the other 5x-6=4x+2 -4x -4x __________ x-6=2 +6 +6 _________ x=8

it's multiply so... (px)y so px+y rule: x^n *x^m = x^m+n

sigh* your right Dr.Bob..shame on me

It's the post right under your last one, not the very last post.

Okay then, I thought that looked kinda funny =)

(3-x)/-4 = (x+6)/3 subtract the terms on the right over to the left (everything) [(3-x)/-4]-[(x+6)/3]=0 find a common denominator for 4 and 3 which is 12 [-3(3-x)/12]-[4(x+6)/12]=0 multiply over (-9+3x)/12 - (4x+24)/12= 0 [(-9+3x)-(4x+24)]/12 = 0 don't forget the negative (-3+3x-4x-24)/ 12 = 0 combine like terms on the numerator then solve for x (-3+3x-4x-24)= 0

Typo after multiply over: Correction is it's -9 not negative -3 [(-9+3x)-(4x+24)]/12 = 0 don't forget the negative (-9+3x-4x-24)/ 12 = 0 combine like terms on the numerator then solve for x (-9+3x-4x-24)= 0

f(x)=lnx f'(x)= 1/x If I remember correctly your going to have to derize the second one again (chain rule) (7x-2)^3 x= 7x-2 dx= 7 (x)^3dx get the derivative of this 3(x)^3-1 dx + c 3(x)^2 dx + c and plugging in the found values... 3(7x-2)^2 ( 7)= 21(7x-2)^2 ---(2nd part) For the first part it Should just be ln(x)= 1/x if I'm not incorrect (my text uses 2 functions instead of using ln so I'm not 100% sure) so putting it together assuming my thinking is correct: (1/x )(21(7x-2)^2) since the 2nd was already differentiated.. by the product rule if I remember correctly f'(x)= (1/x)+(21(7x-2)^2)

I forgot a important part..the first part is to use the product rule derivative of the first * second function + first*derivative of the second then doing this again,correcting that error f(x)= ln(x)(7x-2)^3 product rule first then the chain rule for (7x-2)^3 (1/x)(7x-2)^3 + (lnx)3(7x-2)^2 chain rule for the 2nd part x = 7x-2 dx = 7 ~you could replace the internal equation 7x-2 with x or not but if you do (1/x)(7x-2)^3 + (lnx)3(x)^2dx plug in the values of x and dx and (1/x)(7x-2)^3 + (lnx)3(7x-2)^2(7)= (1/x)(7x-2)^3 + 21 (lnx)(7x-2)^2 (I didn't go and simplify though)

first of all do you have a graphing calculator? Well using my TI-83 and keying in the equations I got (1,3) at the intersection of the 2 lines but for the y intercept it was aproximately (0,6)

just delete this please =)

Oh..so that's why the page went totally blank for a second there alright then, Leo =D

If you want them to equal 17 (5x8-6)/2= 17

since it's a strong acid all of the strong acid is converted to [H+] or [H3O+] so pH= -log[H+]= -log [0.36]= 0,= 0.44

typo correction (highlighted answer) since it's a strong acid all of the strong acid is converted to [H+] or [H3O+] so pH= -log[H+]= -log [0.36]= 0.44

The bile digests fats. (bile is stored in the gallbladder) Bile goes from liver to gallbladder then to the small intestine NOT stomach. (if you had it in your stomach it would be caused by Bile reflux disease) Pancreas info is correct. Both bile and enzymes from the pancreas go to the small intestine. Large intestine absorbs water. (you shouldn't say it's released into the bloodstream but how I said it) Waste then enters the colon where both water is absorbed. then rectum and then it's eliminated.

Both bile and enzymes from the pancreas go to the small intestine. note my quick type error: bile is from the liver then is released into the small intestine from the gallbladder bile is NOT from the pancreas.

Re typing since this is a double post... table of contents type table of contents into Wikipedia (you can type Wikipedia into google search then click that and type table of contents into the search on the left) (I can't link you there unfortunately)

The toungue goes and rolls the food into balls. (bolus) Saliva moistens food and also contains amalayse enzymes which breaks down starches.

tongue not toungue

Your welcome =D Don't worry about it, if you need more help just post and there are many people here including me who are willing to help out if you need it.

Bile is Not an enzyme Alley. It's an alkaline fluid to be exact. other than that: First food goes into the mouth by your teeth in mouth and lubricated by the tongue and the salivary glands secretes saliva in the mouth to help moistening it then it goes to the esophagus where enzymes are secreted on it.split this previous sentence into a few sentences because it's a run on sentence Saliva moistens food and also contains an amylase enzyme which breaks down starches. From here it goes down your esophagus to your stomach. In your stomach your food is broken down into molecules the body can use. The liver produces an enzyme called bile, which helps break down a component in the food you ate. Bile emulsifies fats so they won’t clump in your intestines. From the liver the bile travels to the gallbladder, and then to the small intestines. The pancreas then releases enzymes that break down starches and carbohydrates into the stomach.reword this because you make it sound as if the starches and carbohydrates become the stomach and also in addition, the pancreas enzymes are released into the small intestine not stomach Once your food is broken down into molecules the body can use, it travels through the small intestine where the molecules are absorbed into the bloodstream, and then the rest travels through the large intestine where water is absorbed. Waste then enters the colon.The colon also absorbs water just like I said before The waste enters the rectum where it is pushed into a solid form and eliminated from the body. Other than those corrections it is better than the original.

If that is 2/5 (2/5)/21 = since 21 is actually 21/1 flip and multiply (2/5)*(1/21)= 2/105= 0.01904 ___________________________________ If it's (12/5)/21= 0.1143 Please be more specific as to whether that is a #1 or 1 is part of the question because you put a space in between the 1 and 2/5 which would lead to different answers but not 15.

Then again it could have been a fraction with 1 2/5 = 7/5 if this is true... (7/5)/21= 0.0666 Writing it differently can mean alot of things you see.

I really don't know if you copied the question wrong or something because when I did the question 3 times total and keyed the fractions into my calculator I get the same answer of 485.45 so I don't know if either the multiple choice is incorrect or the copied question is wrong but something is funny.

I just copied it as a decimal (I didn't copy all the 6's) but if I had used the fraction function I would have also gotten 1/15 which is the same.

I'm not sure but a foolproof way to go and use a denominator common to both is to go and multiply both numbers here: 25x47=1175

Brie the 2 fractions multiplied is NOT what you got...are you using a calculator?

Posting my work on this: 16 20/21*28 3/7 - 4 13/25 + 8 2/47 = (356/21)*(199/7)-(133/25)+(378/47)= (70844/147)-(133/25)+(378/47)= 484.654 (this is most likely the answer to this question that you posted [it may have been copied wrong] because I'm using my TI-83 Plus graphing calculator to just straight punch in the fractions(it can handle large fractions) while before I was using a normal calculator) note it's not there as a choice so I personally feel that it's a problem with the question/ multiple choice

Domesticated fish : siamese fighting fish Wild fish: swordfish

since 15miles/60min= 1mile/x min x=4 min 60min= 1hr

range: difference between the highest and lowest value of a data set mean: average of the #'s in a data set mode: # that occurs most often in a data set (the one that repeats the most) median: the middle value of a set of data that is ordered from lowest to highest. (numbers- ex. 123 median is 2) but if the number of data is odd- (ex 1234 the median is an average of the 2 middle #'s (2+3)/2= 2.5

glass containers...

unit rate? well usually it is a unit compared to a rate ex. min/sec= minutes per sec miles/hr= miles per hour

to elaborate: usually corrosives are stored in air permeable containers and then those are stored in storage lockers. Acids and bases are stored separately from each other and other chemicals to prevent rxns with each other. The containers that hold the acid/bases are stored in secondary containers that are chemically resistant and unbreakable too.

To simplify S= 2brh + 2br2 take out the multiple common S= 2br(h+2br) S/2br= h+ 2br (S/2br)-2br= h

Um I forgot to go and put a bracket around the 2br to give (2br)2 since the answer would be different if that was the case but I'm assuming that the one I type now is what it was.

Please repost under a new post on the top of the page emily.

The one you don't test on. All the variables except for the experimental one are used on the control group.

Was what I posted incorrect Dr.Bob?

Visuals always worked for me during that age but in schools they expect us to learn from just words and descriptions alone. I just wished they taught that way all the time. You could go and use a ruler at first to demonstrate the centimeter and inches, a meterstick could help with the meter. Then you could compare each and build from that's such as this many of one measurement equals another unit of measurement. I don't know what you mean by a foo however unless that's a typo.

Ice table ? I haven't heard of that term in awhile. Well since acetic acid is a weak acid and NaOH is a strong base... 120ml 0.15M acetic acid 30ml 0.2M NaOH -You need Ka for the weak acid. (for acetic acid Ka= 1.8x10^-5) steps: 1. Determine if moles H+ or moles OH- is in excess 2. Determine ammount of excess. 3. Determine Molarity of excess ammount. 4. Solve for pH directly or by pOH id OH- is in excess