Saturday
October 25, 2014

Posts by DrBob222


Total # Posts: 44,714

Chemistry
This is a limiting regent (LR) problem. You know that when amounts are given for BOTH reactants. mols AgNO3 = M x L = ? mols K2C4O4 = M x L = ? Using the coefficients in the balanced equation convert mols AgNO3 to mols Ag2CrO4 Do the same for mols K2CrO4 to mols Ag2CrO4 It is ...
October 2, 2014

Chemistry
I agree
October 2, 2014

Chemistry
mols CCl4 = grams/molar mass = ? mols SbF3 = grams/molar mass = ? Convert mols CCl4 to mols SbF3. If the SbF3 mols is greater than CCl4 is the limiting reagent. If not SbF3 is the LR. Post your work if you get stuck.
October 2, 2014

Chemistry
[mass ice x heat fusion]+[mass melted ice x specific heat ice x (Tfinal-Tinitial)] + [mass warm water x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute the numbers and solve for Tf.
October 2, 2014

chemistry
I've ignored the numbers since they seem to be in stoichiometric quantities and no excess of either reagent is left. magnesium metal plus hydrochloric acid yields magnesium chloride + hydrogen. Mg + HCl ==> MgCl2 + H2 Mg + 2HCl ==> MgCl2 + H2 Mg + 2HCl ==> H2 + MgCl2
October 2, 2014

Chemistry
C3H8 is propane, right. C3H6O is not acetone propane. What I drew for you is propene-1-ol
October 2, 2014

Chemistry
Would that be CH3CH=CHOH? 1 double 6 single no lone pairs. Check that.
October 2, 2014

chemistry, biotechnology
If molar extinction coefficient is 1.4 A units/mg then wouldn't c come out in mg?. But that confuses me too. The usual unit is M which is mols/L and this unit you quote is just mg. Yes you should make a correction for dilution. So whatever c is the original solution will ...
October 2, 2014

chemistry
A is right. C is right. I don't agree with B. If dE = q + w and q is zero as you answered in C and work is done as you answered in A, how can dE be zero as you answered in B.
October 2, 2014

Chemistry
gain of heat on melting ice + gain of heat on raising T from 0C to Tfinal + heat lost by 120 g H2O at 70.5C to Tfinal = 0 Each [ ] gives the three processes above. [mass ice x heat fusin] + [mass melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass warm water x specific...
October 2, 2014

Math - Please Check My Answer - Probably Last One!
D is correct.
October 2, 2014

chemistry
1. It would help if we knew what you were recrystallizing. 2. This question makes no sense at all. First they don't melt then they do.
October 2, 2014

chemistry
A = ebc 0.427 = 6130*1*c c in the 10 mL flask = ? M. c in c in 5.00 mL flask = 10x that = ? mols 5.00 flask = M x L g 5.00 flask = mols x molar mass = ? mass% in sample = (g in 5.00 flask/mass sample)*100 =? mass% =
October 2, 2014

chemistry
What's your problem here Mathew? You calculate mols H^+ and mols OH and see where you stand.
October 2, 2014

ph
pH = -log(H^+) BUT I don't see that this has anything to do with your post.
October 2, 2014

chemistry
No it isn't. Go back and look at that work I did for you last night. You have smatterings right but you've incorrectly interpreted what I wrote in several places. mols HCl = 53/36.5 = ? mols NaOH = same as mols HCl because of the 1:1 ratio. g NaOH = mols NaOH x molar ...
October 2, 2014

chemistry
10% HCl means 10g HCl/100 mL so you spilled 10 x 530/100 = 53 g mols HCl = grams/molar mass = approx 1,5 but you need a more accurate answer than that. NaOH + HCl ==> NaCl + H2O You have approx 1.5 mols HCl so you ned approx 1.5 mols NaOH. mols NaOH = gram/molar mass. You ...
October 2, 2014

Chemistry 104
Mg^2+ + NH3 + 7H2O + PO4^3- ==> MgNH4PO4.6H2O + OH^- mols PO4^3- = M x L = 1.0 x 0.010 = 0.010 mols MgNH4PO4.6H2O = 0.010 g MgNH4PO4.6H2O = mols x molar mass = 0.010 x 245.41 = ? This is the theoretical yield. %yield = (2.2034/?)*100 = ???
October 2, 2014

chem
No. As far as I know the Lewis Structure must be drawn. I would look at IO3^- as the culprit.
October 2, 2014

Chenistry
For this problem you do not need pKa1.(Note that pK1 has H3PO4 in it and that isn't one of the ingredients.) I obtained the same answer and I did it with mols ONLY (which doesn't involve dilution and I like that way) but I also did it with molarity and took into ...
October 2, 2014

Chemistry 30
And your problem is?? For a). HNO2(aq) + H2O(l) ↔ H3O^+(aq) + NO2^-(aq) Now you look at thwich has more H and which has fewer. HNO2 is the acid NO2^- on the right is the conjugate base. H2O on the left is the base. H3O^+ on the right is the conjugate acid.
October 2, 2014

Chemistry 30
2NaOH + H2SO4 ==> Na2SO4 + 2H2O mols NaOH = M x L = approx 0.01 mols H2SO4 = 1/2 that from the coefficients in the balanced equation. M H2SO4 = mols H2SO4/L H2SO4 = ?
October 2, 2014

Chem 400
.........I2 + Cl2 ==> IxCly + xs Cl2 .....0.678g..0.851g....0..........x 2Na + Cl2 --> 2NaCl You know NaCl weighed 0.450. Convert that to grams Cl2 and that is the excess amount. Subtract that from the 0.851 g there initially. Now you have 0.678 g I2 and I obtained 0.578...
October 1, 2014

chemistry
CuO + H2SO4 ==> CuSO4 + H2O You want 0.025 mols CuSO4 so you must have start with 0.025 mols H2SO4 (from the coefficients in the balanced equation). Then M H2SO4 = mols H2SO4/L H2SO4. You know M and mols, solve for L and convert to mL. I have no idea what "allowing for...
October 1, 2014

chem
I can't help here but I hope you received my correction last night on the 2-ethylhexene becoming 2-methylheptane.
October 1, 2014

Chemistry
Fe + 2HCl ==> FeCl2 + H2 mols Fe = grams/molar mass = approx 0,06 but you need a better answer than that. Convert to mols HCl which will be twice mols Fe and you get that from the coefficients in the balanced equation. Then M HCl = mols HCl/L HCl. You have M and mols, solve...
October 1, 2014

Chemimstry
So you have approx 0.018 mols Br2. That means twice that for mols XBr so 0.036. Since mol = g/molar mass then molar mass = g/mol = 5.20/0.036 = approx 144. Subtract atomic mass Br from that to find the atomic mass of X and identify from the periodic table. I would look in the ...
October 1, 2014

Chem
g CO2 = 3.626 g H2O = 0.9897 Plan. 1. Convert g CO2 to g C. 2. Convert g H2O to g H. 3. Find g O in the C,H,O compound from 3.297g total - g C - g H = g O. mols CO2 = g/molar mass = 0.0842 and that x 12 = 0.9889 g C g H = 0.9897*2/18 = 0.1099 g O = 2.198 Convert g each to mols...
October 1, 2014

Chemistry
How much hexane was burned? How much O2 was used?
October 1, 2014

chemistry
0.88 what? g/mL; lb/cubic foot; kg/L; tons/gallon? I'll assume you mean 0.88 g/mL. 0.75 lb x 453.6 g/lb = z grams Then 0.88 g/mL x ?mL = z grams from above. Solve for ?mL.
October 1, 2014

Chemistry/Math
It would be great if you would write the entire problem and let us help. Your 0.01403 g doesn't tell us what? Is that the mass CaCO3 or the mass of the sample. And somehow H2O got into the problem. Do you want ppm CaCO3 dissolved in the water?
October 1, 2014

Chemistry/Math
Anyone pick a number and be right.
October 1, 2014

Science
If you mean a changing definition of a CHEMICAL element I must say I didn't know it had changed. I've been using the same definition for the last 70 or more years. I assume you do not mean element as in an element of story or a heating element or something like that.
October 1, 2014

chemistry
Dissolve in water, filter, the AgNO3 is in the aqueous layer, the AgCl is on the filter paper, evaporate the solvent to recover the AgNO3.
October 1, 2014

matter and energy
You didn't provide what you used for the melting point or the heat fusion or the specific heat so I don't know it that is right or not. Using 327.5 for m.p. and 22.4 J/g for heat fusion and 0.16 J/g*C, I obtained 350 Joules.
October 1, 2014

matter and energy
Note the correct spelling of celsius. The melting point of Pb is 327.5 C. q added to raise temperature from 30 C to 327.5C is q1 = mass Pb x specific heat Pb x delta T. You will need to look up the specific heat Pb. delta T is 327.5-30. q2 = heat needed to melt Pb at 327.5 C ...
October 1, 2014

chemistry
mols KCl = grams/molar mass = approx 0.32 g but you need a more exact number for this and all of the estimates that follow. M KCl = 0.32/0.5L = about 0.64 Then mL x M = mL x M 27.3 x 0.64 = 45.0 x M AgNO3 Solve for M AgNO3. The mL x M = mL x M works because the equation is 1:1...
October 1, 2014

CHM 300
pKa = -log Ka
October 1, 2014

chemistry
I don't see a question here.
October 1, 2014

Science
True, the nitride ion is N^3-. False, the Stock system shows the charge on the positive ion.
October 1, 2014

Chem (general)
Your value of 2.19 g Ag2S appears to be right. If you can calculate mols Ag2S why not use the same reasoning to calculate mols H2S used. 2AgNO3 + H2S ==> Ag2S + 2HNO3 mols AgNO3 = 3.00/170 = approx 0.00833 Convert mols AgNO3 used to mols H2S used the same way. 0.00833 mols ...
October 1, 2014

Chemistry
See your later post above.
October 1, 2014

chemistry
HNO3 + xBOH ==> xH2O + B(NO3)x millimols HNO3 = 12 mL x 0.150M = 1.8 millimols base = 30 x 0.03 = 0.9. The titration step shows us that we take twice as much HNO3 as the base; therefore, the base must contain two OH groups; therefore, I would look at Mg(OH)2, Ba(OH)2, Sr(OH...
October 1, 2014

chemistry
HAc + NaOH ==> NaAc + H2O mols NaOH = M x L = estimated 0.008 but you need a more accurate answer for this and all of the other estimates that follow. From the equation. 0.008 mols NaOH = 0.008 mols HAc M = mols/L = 0.008/0.010L = about 0.8M g HAc = mols x molar mass = ...
September 30, 2014

Chemistry
When you say you're given the peaks I assume you mean the frequency or wavelength of each peak. Then you get a book that gives the peaks by element. Peaks for elements different in number, intensity, etc but generally there are 2-5 strong "lines"(peaks) in the ...
September 30, 2014

solution making/ chemistry/biotech
First determine how many mols you have. The molar mass is about 100 so 340g/100 = 3.40 mols. Then mols = M x L. You know mols and M, solve for L That wll be 3.40 = 2M x L M = 3.40/2 - 1.20L or 1200 mL. Yes, you start by 340 x (1/100) = ? which I've done above.
September 30, 2014

Chemistry
A limiting reagent (LR) problem. You know that when you are given amounts for BOTH reactants. mols C = 20/12 = approx 1.7 but you need a more accurate answer for this and all of the estimates that follow. mols Si = g/atomic mass = 65/60 = about 1 mols SiC formed if you had 1.7...
September 30, 2014

Science 1 quick question
Yes, IF they needed N in a form other than N2 gas.
September 30, 2014

Science 1 quick question
If they needed nitrogen they would die without finding some other pathway
September 30, 2014

chemistry
28Si = 92.23% and 27.9769 amu 29Si = X% and 28.9865 amu 30Si = 100%-92.23%-X% = 7.77%-X and 29.9838 amu ----------------------------------- (0.9223*27.9769)+(X)(28.9865)+(0.0777)(29.9838 = 28.0955 Solve for X and 7.77-X
September 30, 2014

oops--Chemistry II
I have a 5-valent carbon below which i need to get rid of. This should take care of it. CH2 || C-C2H5 | CH2 | CH2 | CH2 | CH3
September 30, 2014

chem
This is about the best I can draw it on this forum but I think you get the idea CH2 || HC-C2H5 | CH2 | CH2 | CH2 | CH3 or you can view this site. http://www.chemspider.com/Chemical-Structure.14663.html
September 30, 2014

oops--Chemistry II
I have a problem with the answer. Most is right but the equation should be K2SO4(aq) + BaBr2(aq) ==> 2KBr(aq) + BaSO4(s). Note: KBr is soluble; BaSO4 is not. Next question is how do you know when a solid is formed. The answer is you memorize a set of solubility tables. Here...
September 30, 2014

chemistry
Mg^2+ + Na2CO3 ==> MgCO3 + 2Na^+ Convert mols Na2CO3 to mols Mg^2+ using the coefficients in the balanced equation. mol Na2CO3 x (1 mol Na2CO3/1 mol Mg^2+) = mols Na2CO3 x 1/1 = ? Then grams Mg = mols Mg x atomic mass Mg.
September 30, 2014

Chemistry
I don't see a question here.
September 30, 2014

chemistry
So 100 g/m^3 = 100 g/1000 L = 0.100 g/L. No2 convert 0.100 g phenol to mols. mols = grams/molar mass
September 30, 2014

college chemistry
heat lost by metal + heat gained by water = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] | [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Plug in the numbers and solve for Tf. That's the only unknown.
September 30, 2014

Chemistry
What's the molarity of the H2SO4 you have? That's 1.83 g/mL x 1000 mL x 0.85 x (1 mol/98g) = approx 16 M but you need a more accurate answer. Then c1v1 = c2v2 16M*v = 1.35M x 1000mL Solve for v in mL but remember to get a better answer for that 16M.
September 30, 2014

Chemistry
1. 3O2 --> 2O3 I would call this a synthesis reaction. 2. oxidation reduction 3. synthesis 4. synthesis
September 30, 2014

chemistry 101
Sorry about that. That forms a heptane and the name is 3-methylheptane. You can see that if you draw it out.
September 30, 2014

chemistry 101
2-ethylhexane
September 30, 2014

Chemistry
mols H3PO4 = grams/molar mass Using the coefficients in the balanced equation, convert mols H3PO4 to mols HCl. Now convert mols HCl to grams. That's g = mol HCl x molar mass HCl
September 30, 2014

chemistry
My balanced equation is 4NaClO + 2H2O ==> 4NaOH + 2Cl2 + O2 which is just twice your balanced equation but the ratios are the same. 1.12 g/mL x 15 mL x 0.0585 = 0.982 g NaClO mols = 0.982/74.45 = 0.0132 mols NaClO mols O2 = 0.0132/4 = 0.00330 Then V = nRT/P. I used P = 755/...
September 30, 2014

chemistry
Damon didn't quite balance the equation. There are 3 O atoms on the left and four on the right. What was your balanced equation? I'll look at this later.
September 30, 2014

chem
mols Br in 7.20 mol FeBr2 = 7.2 x 2 = 14.40 Then g Br = mols x atomic mass = 14.40 x 79.9 = ?
September 30, 2014

CHEM
These problems are two equations in two unknowns that are solved simultaneously. Let X = mass As4S4 and Y = mass As4S6 ------------------------ equation 1 is X + Y = 1.000 The second equation comes from the 0.905 g of the As4O6 product formed. The equations are As4S4 + 7O2 ==&...
September 30, 2014

Chemistry
dHrxn = (n*dH BE reactants) - (n*dH BE products) BE = bond energy which you have listed. To get you started C-H is 99 kcal and you have four of them so you must break 4 C-H bonds at 99 each so it costs almost 400 kcal to break those four bonds. Then you put together CO2 and ...
September 30, 2014

Chemistry
IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O Then you titrate the liberated I2 with the thiosulfate. 2S2O3^2- + I2 ==> 2I^- + S4O6^2- How much KIO3 did you weigh out? It was supposed to be weighed accurately but weigh out about 1 g. I will assume you weighed out exactly 1.00 g ...
September 30, 2014

CHEM HELP!
I believe I will do the trick. I don't think 2 will work; flowing the gases through the converter faster will result in less time for reaction and that will be a negative effect. If I looked up the 1/2N2 + 1/2O2 --> NO right ( you should confirm this) it was + meaning ...
September 30, 2014

Chemistry 30
(H2O) = 2.5 mol/50 L = 0.05 M ..........C + H2O ==> CO + H2 I.......solid.0.05.....0....0 C.......solid -x.......x....x E......solid.0.05-x....x....x Kc = (CO)(H2)/(H2O) The problem give you H2, CO is the same H2, you plug in Kc and solve for (H2O) You didn't give Kc in...
September 30, 2014

Chemistry 30
In every day language, Le Chatelier's Principle says that an equation in equilibrium will try to undo what we do to it. Adding O2 means it will shift so as to use up O2. So it goes to the right. Reducing H2O means it will try to make more H2O so it will shift to the right...
September 30, 2014

Chemistry- Answer Check
I wouldn't bet on it. When only one molecule in a thousand has enough energy to react the odds are against much of a reaction at all. I think the question is asking what can you do to make more molecules react and I believe the answer is to raise the temperature. That ...
September 29, 2014

Chemistry
So at 11 C the solubility is 5.6g/16cc. Convert that to g/100 cc and you have 5.6 x 100/16 = ? g/100
September 29, 2014

chemestry
You might check out the spelling of chemistry.
September 29, 2014

chemestry
3Li = 1s2 2s1. If you lose the outer electron it leaves 1s2 which is not an octet because there are only two electrons instead of 8 but it's a closed shell none-the-less.
September 29, 2014

oops---chemistry
1.00 cm x 4.00 cm x 7.80 cm is not 54.7 cc is it? more like 31.2 cc.
September 29, 2014

Chemistry
Bitterness in beer is caused by isohumolones (see http://en.wikipedia.org/wiki/Isohumulone). I couldn't find anything on yeast as far as a chemical structure goes. They are microorganisms, of course. Probably you could use either but since the former has so many different ...
September 29, 2014

Chemistry - Titrations
For whatever it's worth I learned something tonight that I wasn't aware of. It makes no difference how many H ions are neutralized. When mL and M are boxed in with the KOH and NH4OH, the volume comes out to be the same number no matter which assumption one makes. When ...
September 29, 2014

Chemistry - Titrations
15.61 mL should be right. By the way, I should point out that NH4OH doesn't exist. Years ago we thought NH3 + H2O ==> NH4OH which then ionized into NH4^+ and OH^- and that made a lot of sense.(and the ions DO exist.) But years of looking for the NH4OH MOLECULE has ...
September 29, 2014

Chemistry - Titrations
I was waiting for your reply to suggest that the best thing to do is just that; i.e., assume all of three H^+ are neutralized. If that is so we have this equation. H3PO4 + 3KOH ==> K3PO4 + 3H2O millimols KOH = mL x M = 35.25 x 0.2300 = ? mmols H3PO4 used = 1/3 of that = x ...
September 29, 2014

Chemistry - Titrations
M stands for molarity and is = mols/L solution N stands for normality and is = equivalents/L of solution. Most schools have stopped teaching normality and I applaud your teacher for including it. I think it's a huge mistake to dismiss normality BUT the IUPAC has spoken, ...
September 29, 2014

Chemistrey: Please check my answer
Everything is ok except for the last step. The question is for % ClO2 and you substituted mass Cl2 and not ClO2. That should be (25.9/26.98)*100 = ? and round to 3 significant figures. Here is a short cut to calculating mols ClO2. You used ratio/proportion and that will work ...
September 29, 2014

CHM
EDTA complexes with almost all elements. By controlling the pH one can control possible interference from other metals than the one being determined.
September 29, 2014

CHM
It isn't necessary to change screen names for different questions. I have most often heard the term free calcium used in connection with the amount of calcium in solution (but not complexed). I have never heard it used in connection to the amount of Ca metal. For example, ...
September 29, 2014

Chemistry
This is a problem with two equations and two unknowns and you solve them simultaneously. But first you didn't carry out the calculation for mols H2; you threw away some. I solved for n and obtained 0.00719879 which I would round to 0.0072 mols H2. Let X = mass Fe and Y = ...
September 29, 2014

chemistry
This is a problem with two equations and two unknowns and they are solved simultaneously. Let X = mass Ag and Y = mass Cu ----------------- equation 1 is X + Y = 3.00 To get the second equation note that mols NO3^- = M x L = 0.650 M x 0.1L = 0.0650 mols. Therefore, mols NO3^- ...
September 29, 2014

chemistry
What's a ff reaction? The driving force is the formation of I2 which is a solid in effect a precipitate.
September 29, 2014

CHM LAB
Are you doing it with EDTA? mols EDTA = M x L mols CaCO3 = mols EDTA mass CaCO3 = grams = mols CaCO3 x molar mass CaCO3
September 29, 2014

possible oops--Chemistry II
I'm not sure about this and I think the best thing to do is to ignore my response until you've cleaned up the problem so I'm sure of the data.
September 29, 2014

Chemistry II
You need to clean up the post. I know you want normal boiling point of PCl3. Is that dHofF S? Is that -288.07 dHo formation PCl3? Is that 311.7 dSo formation PCl3. dGo = dHo-TdSo dGo at boiling point is zero. 0 = dHo - TdSo Plug in dHo and dSo and solve for T if I've ...
September 29, 2014

Chemistry
Both of those formulas are correct for citric acid. They are empirical formulas and you will notice that both are C6H8O no matter how you slice it. The difference. If you are working a problem and citric acid is acting as an acid you want to highlight those H atoms (H ion) ...
September 29, 2014

Organic
High density sinks. Low density floats.
September 29, 2014

Chemistry
2Na3PO4 + 3Pb(NO3)2 ==> 6NaNO3 + Pb3(PO4)2 mols Na3PO4 = M x L Using the coefficients in the balanced equation, convert mols Na3PO4 to mols Pb3(PO4)2. Then convert mols ppt to grams. g = mols x molar mass.
September 28, 2014

Chemistry
See your post below.
September 28, 2014

Chemistyr
mols Ba(OH)2 = grams/molar mass = approx 0.3 but you need to be more accurate. M Ba(OH)2 = about 0.3/1.20L = approx 0.31 or so. Then use c1v1 = c2v2 0.31 x v = 0.100 x 1.00 Solve for v.
September 28, 2014

Chemistry
https://answers.yahoo.com/question/index?qid=20061013095505AAe2Uvf
September 28, 2014

Chemistry
2Mg + O2 ==> 2MgO mols MgO = grams/molar mass = approx 0.25 but you need a more accurate answer. Then convert mols MgO to mols O2 using the coefficients in the balanced equation. That is 0.25 mols MgO x (1 mol O2/2 mols MgO) = 0.25 x 1/2 = 0.125 mols O2. g O2 = mols O2 x ...
September 28, 2014

chemistry
mols KOH = grams/molar mass = 18.3/approx 56 = approx 0.33 but you need a more accurate number. Then M = mols/L solution. You know M and mols, solve for L.
September 28, 2014

chem
2-ethylhexane
September 28, 2014

Chemistry
Note the correct spelling of celsius. NH4NO3 --->N2O + 2H2O Use PV = nRT and solve for n = number mols N2O at the conditions listed. Use the coefficients in the balanced equation to convert mols N2O from the previous calculation to mols NH4NO3. Then convert mols NH4NO3 to ...
September 28, 2014

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