Friday
October 24, 2014

Posts by DrBob222


Total # Posts: 44,702

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

chemistry
I think I would try converting concn and time to change in rate and plot change in rate vs concn.
July 26, 2014

Chemistry Please Help!!!
http://chem-net.blogspot.com/2011/11/lewis-electron-dot-structures-simple_16.html
July 25, 2014

Chemistry Please Help!!!
This looks like it's down your alley. https://answers.yahoo.com/question/index?qid=20100302133652AAnVmL6 We can't draw structures on this forum. If I can find a lewis dot sturcture the nitride ion I will post as another link.
July 25, 2014

chemistry
mols HCl = M x L = ? mols NaOH = mols HCl M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L and convert to mL.
July 24, 2014

chemistry
Surely you have a graph, a chart, a table, that gives you the solubility of these compounds at this temperature.
July 24, 2014

Chemistry
right.
July 24, 2014

Chemistry
mols KOH = M x L = ? mols H3PO4 = 1/3 mols KOH (look at the coefficients in the equation). M H3PO4 = mols H3PO4/L H3PO4.
July 23, 2014

chemistry
Use the Arrhenius equation.
July 23, 2014

Chemistry
Yes, you need M S2O3^- but surely you have more places than that in the answer.And I wouldn't think 14 mL is close enough. I would think something like four places(that is 14.22 mL). Also I think you have a typo; I believe the first equation after Ksp is not right.That ...
July 23, 2014

Chemistry
How much of this can you do on your own? Where are you stuck?
July 22, 2014

Chemistry
In an enthalpy experiment total heat = (mass H2O x specific heat H2O x delta T) + Ccal x (delta T) I may misunderstand the question but I don't think you subtract the heat capacity of the calorimeter.
July 22, 2014

chem
q = mass Cu x heat vaporization
July 22, 2014

chemistry
m = mols/kg solvent 0.375 = mols/0.570 = ? Solve for mols. Then mol = grams/molar mass You have mol and molar mass, solve for grams.
July 22, 2014

CHEMISTRY (PLEASE HELP)
If your blank is larger than the sample by that amount I suspect contamination o the blank.
July 22, 2014

chemistry
2Zn + O2 ==> 2ZnO mols Zn = grams/molar mass = ? Convert mols Zn to mols ZnO Then g ZnO = mols ZnO x molar mass ZnO. This is the theoretical yield (TY). The actual yield (AY) is 242. %yield = (AY/TY)*100 = ?
July 22, 2014

chemistry
Use the Henderson-Hasselbalch equation.
July 21, 2014

chemistry
How do you define a 3:4 solution of KCl?
July 21, 2014

chemistry
1%........4%.........6% |...3.....|.....2....| An easy way to see this which I learned from Mathmate. So you take 2 parts of the 1% and 3 parts of the 6% to make 4%. So for 100 g you take 40g of the 1% and 60g of the 6% to make 100 g of 4%.
July 21, 2014

Chemistry
The usual things like misreading thermometers, invalid calculations, poor lab technique and all of the other errors likely to be made by a student. Mg may have not been pure, student may not have removed the oxide layer from the Mg ribbon, heat flow into or out of the ...
July 21, 2014

Chemistry
Na is right. The others are not. Na2SO3 is a salt and when it dissolves the SO3^2- hydrolyzes. Na2SO3 ==> 2Na^+ + SO3^2- 1.17M.....2*1.17...1.17 .......SO3^2- + HOH --> HSO3^- + OH^- I......1.17..............0........0 C.......-x...............x........x E.....1.17-x...
July 21, 2014

chemistry
And the question is?
July 21, 2014

Chemistry
A complete calculation can be obtained at https://www.google.com/search?q=calculate+M+ions+in+H3PO4&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a&channel=sb and click on the Chapter 7 pdf file document. They use 5.0M H3PO4 and not 0.2 but the process ...
July 21, 2014

chemistry 185
You have two equations. pH = pK2for H3PO4 + log (base)/(acid) 7.4 = pK2 + log b/a solve for b/a and that's one equation. The second one is acid + base = molarity of solution. Since you want 9g/L and that x 12L you want that many grams in the 12L. Convert that to M. Then ...
July 21, 2014

chemistry
This doesn't look too complicated. Where is your trouble? I would choose A and D.
July 21, 2014

Chemistry
Then I have no idea what the question means. The delta Hformation Mg is zero. For anything else with Mg there must be some kind of reaction.
July 21, 2014

Chemistry
I assume you mean Mg to MgO. How will you burn Mg under water?
July 21, 2014

Chemistry
It means weigh out 12g glacial acaetic acetic acid. The density of glacial acetic acid is about 1.05g/mL and the stuff is about 99.5% pure so if you pipet 11.4 mL that will weigh 12g. However, I don't think you will find a pipet with the volume 11.4 mL. If the 12 g doesn&#...
July 21, 2014

Honors Chemistry
q = mc*delta T q = heat = 72,400J m = 752g c = specific heat = ? dT = 9.95 Solve for c.
July 21, 2014

chem
ascorbic acid is C6H8O6. mols ascorbic acid in 100 g is mol = grams/molar mass = 100/176.12 = ? There are 6 C atoms/mol acid, 8 H atoms/mol acid, and 6 O atoms/mol acid. mols C is 6x that. mols H is 8x that. mols O is 6x that.
July 20, 2014

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