Monday
September 1, 2014

Posts by DrBob222


Total # Posts: 43,438

Chemistry
a. This isn't all inclusive but you would want the food all the mice ate to be the same, living conditions (temperature and humidity), exercise, etc. b. http://chemistry.about.com/od/chemistryterminology/a/What-Is-The-Difference-Between-Control-Group-And-Experimental-Group...
August 8, 2014

chemistry
If it took more alkali (28 mL vs 25 mL HNO3), then the acid was more concentrated than the alkali.
August 7, 2014

chemistry
If you performed this experiment why don't you know the color before and after?
August 7, 2014

Chemistry
This is a limiting reagent (LR) problem. 1a. Using the coefficients in the balanced equation, convert mols CH4 to mols of CO2. 1b. Do the same for mols O2 to mols CO2. 1c. It is likely that these two values for CO2 will not be the same; the correct value in LR problems is ...
August 7, 2014

Chemistry
I have interpreted the problem differently from Bob Pursley so my answers differ. First, the pure acid is what molarity? 1.42 x 1000 x 0.70 x (1/63) = 15.8M Then use c1v1 = c2v2 15.8M x 2 mL = 1M x v2 v2 = about 15.8 x 2/1 = about 31.6 mL = total volume v2 which means you add ...
August 7, 2014

Physical chemistry
I think the trouble you are having is knowing what vapor density is. vapor density = molar mass gas/molar mass H2. vapor density N2O4 = molar mass N2O4/molar mass H2 = 92.011/2.016 = 45.64 vapor density NO2 = molar mass NO2/molar mass H2 = 46.0055/2.016 = 22.82 The ratio is 2:...
August 7, 2014

Chemistry
I think you've made two errors in calculation. If I use 420 kPa and 0.5 L volume with T 800K I solve for n = PV/RT and I obtained your value x 10. Then if I use c = n/v that's n/0.5L = Your value divided by 1000. I think your data are good; I think your calculations ...
August 7, 2014

Chemistry
First, forget what I told you. You are titrating KH2PO4 with HCl. KH2PO4 + HCl --> H3PO4 + KCl You notice (which is what I didn't notice when I first read the problem) that 10 mL of 0.1M KH2PO4 is exactly neutralized (neutralized may be the wrong word) by 10 mL of 0.1M ...
August 7, 2014

Chemistry
Use the Henderson-Hasselbalch equation.
August 7, 2014

chemistry -repost
A couple of points. 1. Did you mean "occurs quickly, a flash of light creates an I2 concn ......? so that the rate law is rate = k(I)^2 for 2I ==> I2 2. I have been unable to use this to obtain 1.0E-6 but I have close to that. 3. I will look at it again if you will ...
August 6, 2014

Dance
high kickers hot kickers
August 5, 2014

Chemistry
PV = nRT Solve for n = number of mols, then n = grams/molar mass. You know n and grams, solve for molar mass.
August 5, 2014

Organic Chemistry
Is this to show your answer? I would agree with OH^- to H2O to H3O^+ but I don't know what H3O- is. And I don't understand part b.
August 4, 2014

Science(check answers)
I agree with 2 and 3 but I don't see questions 4,5,6.
August 4, 2014

chemistry
I am far from an organic chemist but I would think it would hydrolyze to form salicylic acid and acetic acid. I don't know if the SA would decompose after that or not.
August 4, 2014

Ochem
%v/v = (volume eugenol/total vol)*100 = 4.42 total volume = 30 mL; therefore, 0.0442 = (volume eugenol/30) volume eugenol = 0.0442*30 = ? but I would have done it in grams from the beginning and used density to convert to mL. (m eug/mH2O) = (peug*164/pH2O*18) (m eug/30) = (4*...
August 4, 2014

Chemistry
See below
August 4, 2014

Chemistry
See your post below.
August 4, 2014

Chemistry
You didn't give any volumes. First, determine the volume for the equivalence point. HN3 + KOH ==> KN3 + H2O mL HN3 x M HN3/mL KOH = mL KOH needed to reach the equivalence point. Then divide the titration into segments as follows: a at the beginning, c at the equivalence...
August 4, 2014

Chemistry
dS = nRT*ln(Vfinal/Vinitial)
August 4, 2014

Chem
P is 155/760 = ? if you are using R = 0.08205 and T is 273+339 = ? That gives you (NO)in mols/L and you plug that into the rate law to calculate [delta(NO)/delta T] and convert that back to mm Hg using P = MRT (remember you will get atm and atm x 760 = mm Hg)
August 4, 2014

Chem
Your post said you didn't know how to derive the formula. I did that for you. Do you have an answer for the problem? I assume so since this is a study guide.
August 4, 2014

Chem
Could this be as follows: PV = nRT with P in atm, V in L, n = mols and R and T usual.' P = n(RT)/V Since n/V = mols/L that is M so p = MRT.
August 4, 2014

chemistry
1.0 x 10^-7 WHAT?
August 4, 2014

Science (Physics)
I = E/R
August 3, 2014

Science
1/R = 1/R1 + 1/R2 + 1/R3 Plug any value into R and calculate it.
August 3, 2014

Physics
I = E/R
August 3, 2014

Physics
See your post above with the 3 ohm and 6 ohm R
August 3, 2014

chemistry
The corrected equation is as follows and I've balanced it as well: 2KNO3 + H2SO4 = K2SO4 + 2HNO3 mols H2SO4 needed = 2 mols KNO3 x (1 mol H2SO4/2 mol KNO3) = 2*1/2 = 1 mol H2SO4 needed. g H2SO4 = mols H2SO4 x molar mass H2SO4. b. 5 mols H2SO4 x (2 mol HNO3/1 mol H2SO4 = 5 ...
August 3, 2014

chemistry
For what reaction?
August 3, 2014

Chemistry
a. Na is the limiting reagent. You should verify that. mols Na = 5/23 = about 0.22 but that's an estimate. [386 kJ/(2*23g)] x 5 = kJ heat released. b. 5 g Na = about 0.22 mols Na and that will produce 0.22/2 = about 0.11 mols H2. Use PV = nRT to solve for V in L. If you ...
August 3, 2014

chem
OK. I will assume the Ksp for CaSO4 is about 1E-4. The values I can find on the web aren't quite that but it makes the problem a little easier. Since you can't find it in your tables, I assume the problem is saying that 1.00 g CaSO4 is soluble in 1 L solution. If that ...
August 3, 2014

chem
Right, however, look in the tables in your text (usually in the back in the appendix) and you will find Ksp (solubility products) listed. Those are the numbers I need. Why can't I look them up? I can but tables don't agree and I want to use the same numbers you must use.
August 3, 2014

chem
What Ksp values are you using for BaSO4 and CaSO4?
August 3, 2014

Chemistry
Not quite. If you use Kb for NH3 = 1.8E-5 then pKb = -log Kb = about 4.74 and since pKa + pKb = pKw = 14, then pKa = 14-4.74 = 9.26. Your tables may give a different value for Kb NH3 but most show 1.8E-5 or 1.75E-5
August 3, 2014

Chemistry
The base is NH3. The acid is NH4^+. a. With a strong acid it's the base that uses it; i.e., NH3 + H^+ ==> NH4^+ b. With a strong base it's the acid that uses it. NH4^+ + OH^- ==> NH3 + H2O c. So we start with 100 mL of the buffer. millimols NH3 = mL x M = 100 x ...
August 3, 2014

Chemistry
Jason, have you worked with the Henderson-Hasselbalch equation?
August 3, 2014

Chemistry
After you go through the equilibrium the second time you end up with M COCl2. You know M = mols/L. You know M from your calculation and you know L (10L) so mols = M x L = ? = n and use PV = nRT to solve for pressure. No, you don't go through the Kp thing because there is ...
August 3, 2014

Chemistry
You solve the equation. mols Cl2 = 2.60/71 = 0.0366 M Cl2 = mols/L = 0.0366/10 = 0.00366 M CO = 0.00366 Kc = (COCl2)/(CO)(Cl2) 1.23E3 = [(x)/(0.00366-x)^2] 1.23E3*(0.00366-x)^2 = x 1.23E3*(0.00366-x)(0.00366-x) = x and go from there. x = 0.0023 I believe although I've ...
August 3, 2014

Chemistry
These numbers are estimates so you need to go through and recalculate all of them. (Cl2) = mols/L = 2.60/71/10 = about 0.004M (CO) = 0.004 (actually closer to 0.00366). ..........CO + Cl2 ==> COCl1 I.....0.004..0.004.......0 C........-x.....-x.......x E.....0.004-x.0.004-x...
August 3, 2014

Chemistry
Reason it out. The ideal gas law is PV = nRT If you keep V, R and T constant, then P is proportional to n so larger n values will increase P.
August 2, 2014

Chemistry- practice question
qrxn is -8537.4 J q/gram = -8537.4/0.47 = ? q/mol = -8537.4/0.47g)*(24.3 g/mol) x (1 kJ/1000 J) = -441.4017 kJ/mol but if that 0.47 g Mg is correct you're allowed only two significant figures.
August 1, 2014

Chemistry
(P1V1/T1) = (P2V2/T2)
August 1, 2014

chemistry
mols compound = grams/molar mass = ? M = mols/L solution = mols/0.142 L = ?M. That x 1000 = mM.
August 1, 2014

science
mols CH4 = grams/molar mass = ? mols O2 = g/molar mass = ? mols SO2 = g/molar mass = ? total mols = mols CH4 + mols O2 + mols SO2 XCH4 = mols CH4/total mols pCH4 = XCH4*Ptotal
July 31, 2014

Chemistry
If you are satisfied that dG is ok, then set up the ICE chart. .......H2O ==> H^+ + OH^- I......1.0.....0......0 C......-p......p......p E......1.0-p...p......p Then Kp = pH^+ * pOH^-/pH2O and substitute the E line into Kp expression and solve for p. Then use p to calculate...
July 31, 2014

science
density = mass/volume
July 31, 2014

Chemistry / help
I really don't understand why you are having trouble with this? It appears to me the answer should be obvious to you. Tell me what you don't understand and we can go through it. But let me give you some food for thought. 1. Can you place a cube of ice (a solid) in a ...
July 31, 2014

chemistry
Lactic acid = HL, sodium lactate = NaL. ..........HL ==> H^+ + L^- I...... 0.125....0.....0 C.........-x.....x.....x E......0.125-x...x.....x Substitute the E line into Ka expression and solve for x = (H^+), the % ion = [((H^+)/(HL)]*100 = ? for b part, plug into Ka ...
July 31, 2014

Chemistry
(H^+) = sqrt(KwKa/Kb), then convert to pH or pH = pKw+pka-pKb
July 31, 2014

chemistry
I would go with c as the false statement.
July 30, 2014

Chemistry
Two picky points here. 1. You have 3 significant figures in 0.100 and three in 15.9 (and I'm calling 250 3 although some would call that only two since there is no decimal after the zero) so I would keep 3 in the answer and not round it to 1.6 (that is about 1.57 or so). 2...
July 30, 2014

chemistry
I suggest you consider posting all of the problem. You won't get NaF from S and F.
July 30, 2014

Chemistry
Ah Ha! So you're posting the problem right. And you see there is a difference in m and M. How many mols do you want? That's M x L = mols. Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
July 30, 2014

Chemistry
Interesting post but I don't see a question.
July 30, 2014

Chemistry
I don't think so. If dS is + then the term -TdS will always be -. So dG will be - if dH is not too + to more than balance out the -TdS term.
July 30, 2014

Chemistry
Add eqn 1 to 1/2 the reverse of eqn 2 + 1/2 eqn 3 to obtain the final equation you want. Then add the delta H values EXCEPT if you take 1/2 the equation also take 1/2 dH and if you reverse an equation then you change the sign of dH.
July 29, 2014

Chemistry
dHrxn = (n*dHf products) - (n*dHf reactants)
July 29, 2014

chemistry
c1v1 = c2v2 1.0M x v1 mL = 0.25M x 100 mL v1 = mL of the 1.0 M stuff. Add to 100 mL volumetric flask and make to the mark with distilled water. Mix thoroughly.
July 29, 2014

Chemistry
2.0 what? m Is that 0.69m or 0.69M? I assume 0.69m That means 0.69 mol/0.250 kg solvent. 0.69 x molar mass Al2(SO4)3 = approx 60g. Then the solvent has a mass of 250 g for a total mass of approx 250+60 = approx 0310 g. What's the density of the solution? volume = mass/...
July 29, 2014

Chem 2
In pure H2O. ........Ca(OH)2 ==> Ca^2+ + 2OH^- I.......solid........0........0 C.......solid........x........2x E.......solid........x........2x Ksp = (Ca^2+)(OH^-)^2 Substitute the E line into Ksp expression and solve for x = solubility in mols/L and convert to g/L. In 0....
July 29, 2014

Chemistry
There is a tremendous amount of typing here. Here is a summary of what you do. Respond by letting me know how many of these you know to do and where your big problem is. ..........HN3 + KOH ==> KN3 + H2O The first thing you need to do is to determine the mL needed to reach ...
July 29, 2014

Chemistry
Vinegar is acetic acid (CH3COOH) but I will shorten that to HAc. 5.0% w/w means 5g HAc in 100 g solution. g HAc in 2.5g vinegar is 0.05 x 2.5 = about 0.125 mols HAc = g/molar mass = ? mols KOH = mols HAc M KOH needed = mols KOH/L KOH. You know mols and M, solve for L and ...
July 29, 2014

Chemistry
If you know the solubility rules (#1) in my explanation, then you know BaSO4 is a solid (I show solid by BaSO4 in the equation I wrote belowB) and you know BaCO3 is a solid (which I show solid in my equation below). So the first two don't react, the next two react to ...
July 28, 2014

Chemistry
Ba(NO3)2 mixed with NaCl NO Ba(NO3)2 mixed with HCl NO Ba(NO3)2 mixed with Na2CO3 YES Ba(NO3)2 mixed with Na3SO4YES but you have a typo in Na2SO4 also Ba(NO3)2 is soluble and a solid and HCl is a gas. I don't know what you're asking here. HCl is a gas but usually you ...
July 28, 2014

chemistry
CaC2 + 2HOH ==> Ca(OH)2 + C2H2 mols C2H2 = grams/molar mass Using the coefficients in the balanced equation, convert mols C2H2 to mols H2O (HOH). Now convert mols H2O to g. g = mols H2O x molar mass H2O = ?
July 28, 2014

chemistry
Just to be picky I think this problem is not stated correctly (although who am I to know what the author was thinking when this problem was written?). However, I think the spirit of the question expects an answer that Steve gave. Let me point out though that the question asks ...
July 28, 2014

Chemistry
0.04% means 0.04g/100 mL (w/v) so in 200 mL that is 0.08g or 80 mg. So 2.5*X = 80 X = ?
July 28, 2014

Science
I agree
July 28, 2014

Chemistry
ether ==> ethanol Balance the equation. mols ether = tonnes/molar mass (OK, this isn't mols but a tonne-mole but we will be consistent throughout so this is ok. If you don't do this you must convert tonnes to grams in this step and from grams back to tonnes in the ...
July 28, 2014

Science
I agree
July 28, 2014

organic chemistry
Sorry but we can't draw diagrams on this forum. You may be able to find it on Google.
July 28, 2014

chemistry
I asked earlier what you meant by a 2:5 and 3:5 solution and you didn't answer. I asked because there are several meanings to this. In my book a 2:5 solution is 40%(200/5) and a 3:5 solution is 60% (300/5). So I use c1v1 = c2v2 40*30 = 60*v2 v2 = 20 That is, add 20 mL of ...
July 28, 2014

Chemistry
Few people memorize solubility tables. Surely you have a graph or table giving the solubility of KNO3 vs temperature. If you will post the solubility in those tables/graph at 80 C and 50 C I can show you how to do this.
July 28, 2014

physical science
H2C2O4 + 2NaOH ==> Na2C2O4 + 2H2O Temperature increase if the M of each is high enough.
July 28, 2014

Chemistry
mols NH3 = 59/molar mass NH3 = ? Convert mols NH3 to mols NO using the coefficients in the balanced equation. That's ?mols NH3 x (4 mols NO/4 mols NH3) = ? x 1/1 = ?mols NO. Then ? mols NO x 22.4L/mol = x L NO @ STP
July 28, 2014

science
g PbSO4 = 1.645 mols PbSO4 = 1.645/molar mass = estimated 0.00542 = estd 0.00542mols Pb = estd 0.00542 mols PbCl2. g PbCl2 = mols PbCl2 x molar mass PbCl2 and that is g/250mL. That x 4 = g/L Then g/L divided by molar mass PbCl2 = M = (PbCl2) = (Pb^2+). (Cl^-) = 2 x (Pb^2+). ...
July 27, 2014

chemistry
mass = volume x density or volume = mass/density Substitute and solve for volume. If you use density in g/mL you must change mass from kg to g. 52.3 kg = 52300 g.
July 27, 2014

chemistry
mols SO2 = 610/64 = estimated 9.3 mols O2 = 280/32 = estd 9 mols SO3 formed = 750/80 = about 9 Note: You must clean up all of the numbers. I've just estimated them here. ...........2SO2 + O2 ==> 2SO3 I..........9.3.....9.......0 C..........-2x....-x.......2x E.........9...
July 27, 2014

chemistry
........CO + H2O ==> CO2 + H2 I.......10....10......0.....0 C........-x...-x......x.....x E.....10-x...10-x.....x.....x The problem tells you that CO2 is 7.4 mols (so x = 7.4); therefore, at equilibrium CO is 10-7.4 = 2.6 = H2O CO2 = H2 = 7.4 Then total mols = you add them...
July 27, 2014

chemistry
............N2O4 ==> 2NO2 E............72......29 Kp = pN2O4/(pNO2)^2 Substitute and solve for Kp.
July 27, 2014

Chemistry
I agree
July 27, 2014

chemistry check answer
Alter the enthalpy of what Everyone knows that dropping solid NaOH pellets into H2O is a HUGE exothermic reaction.
July 27, 2014

chemistry
actual yield (AY)= 65.3 kg calculated = theoretical yield (TY)= 70.6 kg. %yield = (AY/TY)100 = ?
July 27, 2014

chemistry
How much CO2 should you obtain? That's 7.5 mol O2 x (2 mols CO2/1 mol O2) = 7.5 x 2 = 15 mols CO2 = theoretical yield Then %yield = (Actual/theoretical)*100 80 = (Actual/15)*100 Actual = ?
July 27, 2014

chemistry
First, calculate theoretical yield (TY). Zn + 2HCl ==> ZnCl2 + H2 mols Zn = grams/atomic mass = about 0.05 but that's an estimate. Using the coefficients in the balanced equation, convert mols Zn to mols ZnCl2. That's about 0.05 x (1 mol ZnCl2/1 mol Zn) = estd 0.05 ...
July 27, 2014

science
Fe has two electrons to give away; S needs two electrons to complete its outer shell. Bingo. FeS
July 27, 2014

Chemistry
There are several ways to do this and this method I'm showing you may be the longest but it's the first one I came up with. So here goes. This is a two equation problem and you solve the two equations simultaneously. Here are the steps. 1. Set up equations to calculate...
July 26, 2014

CHEMISTRY
Frankly I don't think there is an easy way to do it.
July 26, 2014

chemistry
http://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity
July 26, 2014

wavelength
Here is a very very very loooong answer for you (but it's very well done). You can skip to the last two lines and get your answer but if you read all of it you will summarize a lot of physics/chemistry/wave/particle information. http://www.physlink.com/education/askexperts...
July 26, 2014

Physics
I don't see any brackets but I see parentheses. I don't see any answers. I would answer the first one with "shifts". I would answer the second one with "slits". For #2 it's much much easier to observe interference from light waves by looking at ...
July 26, 2014

chemistry
New M HCl after dilution is 1.20M x (560 mL/6.10 mL) = ? M HCl. Then pH = -log(HCl)
July 26, 2014

Chemistry
So you have 1/2 mol Fe2(C2O4)3 and 1/2 mol FeC2O4. Write the reaction for the ferric salt in which just the C2O4 is oxidized to CO2. Then write the reactiion for the ferrous salt in which ferrous is oxidized to ferric and the oxalate is oxidized to CO2. Then add the mols ...
July 26, 2014

Chemistry
.....M2(SO4)3 + 3BaCl2 ==> 3BaSO4 + 2MCl3 .....0.596g.................1.220g Convert 1.220g BaSO4 to mols. mol = g/molar mass = estimated 0.005 Convert to mols M2(SO4)3. That's estd 0.005 x (1 mol M2(SO4)3/3 mol BaSO4) = 0.005*1/3 = about 0.00174 but you need to sharpen...
July 26, 2014

Chemistry
1575 kg 1575000g 1575000/17 = estimated 9E4 mol All of the NH3 at the beginning produces HNO3 (at 100% yield) so 9E4 mol NH3 will produce 9E4 mol HNO3. That will be 9E4 mols x 22.4 mol/L = estd 2E6 L if everything is 100%. But these are not. The efficiency is 0.5 x 0.6 x 0.8...
July 26, 2014

chemistry
http://en.wikipedia.org/wiki/Binding_energy http://en.wikipedia.org/wiki/Nuclear_binding_energy http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html http://www.chem.purdue.edu/gchelp/howtosolveit/Nuclear/nuclear_binding_energy.htm
July 26, 2014

Chemistry
NaBr + AgNO3 ==> AgBr(s) + NaNO3 mols NaBr = grams/molar mass = estimated 0.25 mols AgNO3 = M x L = estd 0.09 So all of the AgNO3 will be used and you will have 0.25-0.09 = 0.163 mols NaBr unreacted. (NaBr) = (Br^-) = mols/L = 0.163/0.150 =? I have assumed that the amount ...
July 26, 2014

Chemistry
wt Fe2O3 = 0.8 x 1/2 = 0.4gram. Convert 0.4g Fe2O3 to g Fe. That's 0.4g x (2 atomic mass Fe/molar mass Fe2O3) = estimated 0.28g Fe = estd 0.005 mol Fe (that's mol = g/atomic mass). So how much OH^- is needed? Fe^3+ + 3OH^- ==> Fe(OH)3 0.005 mols Fe^3+ will require 3...
July 26, 2014

chemistry
A straight line parallel to x axis is zero order. A straight line between x and y axis passing through zero is 1 st order (like a Beer's Law plot) A curved line is second order.
July 26, 2014

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