# Posts by DrBob222

Total # Posts: 55,882

**Chemistry-kd**

Not to worry. Yes, it is a general equation. If you were to do 4 extractions you would need to go through the Kc = [(mg x org layer/volume)/(mg x aq layer/volume)] four times. After the first time you would take the amount in the aq layer, extract that a second time, take what...

**Chemistry-kD**

I assume when you say 10 mg remained that it was 10 mg in the aqueous layer. 1. f = [1+ (Kd*Vo/Va)]^-n f is the fraction X recovered. Kd is distribution coefficient with organic layer/aqueous layer usually written as Ko/a - Kc. Vo is volume of organic layer. Va is volume of ...

**Chemistry**

yes

**Chemistry**

You have the right idea but the wrong numbers. You have calculated mols Al. You want mols Al2O3; the problem asks for mols Al2O3.

**Chemistry**

I agree with 5.02 g Fe produced.

**physical science**

Actually, they don't. Most liquids have a u shape but liquid mercury has the reverse. If it is a U shape, the liquid adheres to the wall of the tube and that attractions is stronger than the attraction between molecules of the liquid in the middle part of the U.

**Chemistry-titrations**

I assume Hph stands for phenolphthalein, or if not, then it changes color in the same pH range as phenolphthalein which is approx 8.3 or so. Also, I assume that all of the (NH4)2CO3 reacts, all of the NH3 is boiled away, and an excess of NaOH remains with the Na2CO3. Your ...

**chemistry**

Is it more dense or more viscous.

**Chemistry**

If you follow up I suggest you post new at the top; otherwise, this will be so far down the list that it will be difficult to find.

**Chemistry**

From your last post, initial mols Ba(OH)2 = 0.4 x 50E-3 or if you want to use the 0.2, because the 50 cc was diluted to 100 cc, then mols Ba(OH)2 = 0.2 x 100E-3. Either way you have same mols which is why I like to use the initial mols Ba(OH)2 as 0.4*50E-3, work through the ...

**Chemistry**

You don't post easy questions. Thanks for sharing your thoughts. That helps. Here are my thoughts/questions. 1. The problem says make appropriate assumptions and calculate (HA) and Ka. I'm not sure what "appropriate" means so I went through the possibilities...

**Chemistrt**

Let's let lambda be L and drop the n1 and n2. Since n1 = then that squared is 1/4 and n2^2 will just be n^2 1/L = R(1/4 - 1/n^2) 1 = RL(0.25-1/n^2) 1 = RL[(0.25n^2-1)/n^2] n^2 = RL(0.25n^2-1) n^2 = 0.25RLn^2-RL n^2-0.25RLn^2 = RL n^2(1-0.25RL) = RL n^2 = [RL/(1-0.25RL)] ...

**Chemistrt**

If you have a problem with numbers, the easy thing to do is to leave the equation as is, substitute the numbers, the solve for the unknown in the equation. If you get stuck, post the numbers and someone can help you through.

**Physical science CHECK my answers**

5 is not right. The others look OK to me.

**Chemistry**

You didn't provide any data. You need to fill in the blanks.

**Chemistry**

Ethanol is polar and has hydrogen bonds.

**Chemistry**

mols = grams/molar mass so grams = mols x molar mass

**Science**

I don't know how useful this will be but here it is. http://k12.phys.virginia.edu/Labs/Lab05.pdf Also check out the directions, especially 2b 3. That should be a good starting point for an experiment. Good luck.

**@shenaya**

I posted a response to your Ksp/K but I was late doing it. Hope you saw it.

**Chemistry**

mols Ar = grams/molar mass = ? mols Ne = grams/molar mass = ? mols unk = grams/molar mass but you don't know the molar mass of this; therefore, you can't calculate this at this point. total mols = mols Ar + mols Ne + mols unk. Use PV = nRT to calculate total mols. Then...

**Chemistry**

I don't believe the Eo values you have but we'll go with what you posted. Zn ==> Zn^+ 2e Eo = +0.96 Cu^2+ + 2e ==> Cu Eo = +-.3 ----------------------------- Zn + Cu^2+ ==> Zn^2+ + Cu Ecell = 0.96+0.3 = ? Then dG = -nEF Calculate dG. n is 2; F is 96,485 ...

**chemistry(a mystery)**

chlorine. Cl2 is a single bond; N2 is a triple bond. Cl has just 1 electron to add to make an octet and it forms an ionic bond. N needs 3e to make an octet and few N compounds are ionic; i.e., most are covalent.

**Chemistry**

let x = mL of 0.4 and 50-x = mL of 0.025 --------------------- 0.025(50-x) + 0.4x = 50(.04) solve for x and 50-x

**chemistrey**

(p1v1/t1) = (p2v2/t2)

**chemistry**

mols NaH2PO4 = M x L = ? mols Bi^3+ = mols NaH2PO4 from the coefficients in the balanced titration equation (1:1). Convert mols Bi^3+ to mols eulytite by mols Bi^3+ x (molar mass eulytite/2*atomic mass Bi) = ? Then g eulytite = mols eulytite x molar mass eulytite. Then % ...

**chemistry**

mols I2 initially = M x L = ? Excess I2 titrated with S2O3^2- is I2 + 2S2O3^2- ==> 2I^- + S4O6^2- mols S2O3^2- = M x L = ? Convert mols S2O3^2- to mols I2(that will be mols S2O3^2- x 1/2 = ? and this is the excess I2). I2 used in the Pb(C2H5)4 = mols I2 initially - mols I2 ...

**oops---Chemistry-ksp and kb**

I see I wrote y = (M^+) and that isn't right. y = (MOH) = (X^-). You know Kb for MOH and you know OH so you can calculate (M^+).

**Chemistry-ksp and kb**

y gives the concn of X^- and therefore of KX. And X^- is not hydrolyzed for the reason you state. This question is so far down the list that I almost assumed you had not responded to my first response. If you have further comments please copy and post a new at the top of the ...

**Chemistry-ksp and kb**

I had trouble following your thoughts after the first few lines but the first part is OK. You started out right. You need Ksp/Kb = 0.15 ......MX ==> M^+ + X^- Ksp = .... ......OH^- + M^+ ==> MOH 1/Kb =... ------------------------------- Add to get MX(s) + OH^- =>MOH...

**Chemistry**

https://www.google.com/search?q=crysrallization&ie=utf-8&oe=utf-8

**Chemistry**

See your other post. Read the section in Wikipedia.

**Chemistry**

Sublimation first and seeding second. If you had ever done a crystallization you would know the answers to all of these.

**chemistry**

http://www.jiskha.com/display.cgi?id=1480355412

**math**

YOU CANNOT COPY AND PASTE TO THIS SITE!

**math**

You can't copy and paste to this site.

**science**

What is the investigation?

**i need help with mah**

I see no data.

**Chemistry**

mL1 x M2 = mL2 x M2 12 x 0.790 = 50 x M2 OR, since it's dilution, you start with 0.790M and multiply by a factor less than 1 like this 0.790 x 12/50 = ? Technically, M2 is the concentration of Cu(NO3)2 but since 1 mol Cu = 1 mol Cu(NO3)2 it's all the same.

**math**

A has a horse worth $900 initially. When A goes through the various transactions he ends up with the same horse now worth 891. $900-891 = 9.00 I see how you came up with $171. You're confusing loss and gain with the worth of the horse. You're mixing apples and oranges...

**oops--typo--math**

I hit the - key instead of the = key. That line should be The loss to A is 900-891 = 9.00.

**math**

A has the horse at $900. A sells the horse to B at 10% loss so 900 x 0.1 = 90 and 900-90 = $810. Now B sells the horse with a 10% profit or 810 x 0.1 = 81 and 81 + 810 = $891. The loss to A is 900-891 - 9.00. I don't know how you come up with $171. Are you converting the ...

**Math**

I don't think so. Put this is numbers as in the problem with x and y dollars.

**Math**

He makes $50 and saves $40 so he spends $10. Right? Convert that to x and y and you have it. Don't get confused with x and y. Sometimes it helps to put it in numbers.

**Math**

let x = number of seats. Then (2/3)x = 300 Solve for x.

**Math**

let x = total students then (1/12)x = those absent and (11/12)x = those in school But (1/5) of the (11/12)x went on field trip = (11/60)x. So x - (11/60)x - (1/12)x = 704 Solve for x.

**Chemistry**

I'm not sure EXACTLY what you did. 1 is OK 2 is OK 3 is not right. The mass H2O is 139.0 g and q = mc(dT). You didn't use specific hrat H2O as far as I can see. 4. Apparently you are to calculate the initial T of the metal. [mass Cu x specific heat Cu x (Tf-Ti)] + [...

**Chemistry**

Plug and chug.delta T is 2.8

**Chemistry**

You don't ask the question but I assume you will be calculate delta H or something like that. Yes, use q = mcdT. That's right, you don't have Tfinal and Tinitial BUT the problem tells you T increased by 28 C so Tfinal-Tinitial = 28

**chemistry**

http://www.jiskha.com/display.cgi?id=1495566182

**chemistry**

You have typed your entire lab experiment here. I'm not inclined to answer each line by line question but I'll be glad to help if you tell me what I can help you with. Note that you have initial volumes and final reading the buret. Subtract the two to find the volume ...

**Chemistry**

1. Use the solid. How many moles do you want? That's M x L = 0.5 x 0.100 = 0.05 How many grams is that? mols = grams/molar mass or grams = mols x molar mass = 0.05 x 40 g/mol = ? Weigh ? grams NaOH, place in a 100 mL volumetric flask, add some water, swirl to dissolve, add...

**chemistry**

mols Mg = grams Mg/atomic mass Mg = ? Using the coefficients in the balanced equation, convert mols Mg to mols H2. Now convert mols H2 to volume (in L) knowing that 1 mol will occupy 22.4 L.

**Science/chemistry**

You transposed some letters; I assume you meant CO2 and not OC2. mols CH4 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols CH4 to mols H2O. Then convert mols H2O to grams. grams H2O - mols H2O x molar mass H2O Post your work if you get stuck.

**Chemistry**

Use PV = nRT You have P, R, T. n = grams/molar mass. Don't forget to convert 35C to kelvin.

**Math**

Let x = speed of the boat. Then x+7 is total speed with the current. Against the current it is x-7 x+7 = 3(x-7) Solve for x.

**Chemistry**

KHP + KOH ==> K2P + H2O mols KHP = grams/molar mass = ? mols KOH = mols KHP (look at the coefficients and see they are 1:1). Then M KOH = mols KOH/L KOH.

**chem**

......PbI2 ==> Pb^2+ + 2I^= I....solid......0.......0 C....solid......x......2x E....solid......x......2x The other ionization is NaI and that is ionized 100% (completely that is). ........NaI ==> Na^+ + I^- I.....0.05M......0.....0 C....-0.05.....0.05...0.05 E......0...

**chemistry**

The KMnO4 oxidizes the Fe(NH4)2(SO4)2 from iron(II) to iron(III) and is reduced from Mn(VII) to Mn(II) in acid solution

**chemistry- morality**

I don't know about the morality of these answers but I'll try. First, I assume your 8.00 m actually is 8.00 M. m means molality. M means molarity. Use the dilution formula of mL1 x M1 = mL2 x M2 1.19 x 8.00 = mL2 x 1.50 Solve for mL2. Post your work if you get stuck.

**chemistry**

I'm at a loss since I don't know what experiment, the details, etc. If the "final" solution is the KHX, adding NaOH will titrate the H and leave K2X as if more NaOH were added in the first place as in H2X + 2KOH ==> K2X + 2H2O

**chemistry**

Follow along and confirm my conclusions. My rationale is as follows: millimols KOH = 50 x 0.2 = 10; therefore, any H2X < 10 mmols will leave excess (xs) KOH in the solution and that will be more basic. So we must have at least 10 mmols H2X. Using exactly 10 mmols H2X will ...

**Chemistry**

7.5 mg MeOH/mL x 1000 mL = 7500 mg/L or 7.5 g/L. mols MeOH = grams/molar mass and M = mols/L. Then use the dilution formula of mL1 x M1 = mL2 x M2

**Chemistry**

I have copied this question and moved it to the top where I hope it will be easier to find. Please make any comments to that location and not to this one.

**Chemistry**

Now I'm having doubts. As I think about it I'm suspicious that 0.002 just doesn't sound like a number that would be right. And if you look at 50 cc of 0.002 M H2X reacting with 50 cc of 0.2M KOH,it doesn't make sense that there is KOH left over sine an excess ...

**Chemistry**

I think you are VERY close but not quite there. For 1,I think you substituted into Ka1 to solve for C but in reality I think you should have used the Henderson-Hasselbalch equation because you have formed a buffer with KHX and H2X. pH = pKa1 + log (base)/(acid) 7 = 5 + log(0.1...

**Chemistry**

mols HCl = M x L = ? mols NaOH = mols HCl (look at tahe coefficients; the rxn is 1:1). Then M NaOH = mols NaOH/L NaOH.

**chemistry**

I don't believe there is a reaction between NH4Cl and KMnO4.

**Chemistry**

I like to work in millimols = mmols. M = mmols/mL. So you ask yourself how much NaOH it would take to neutralize ALL of the H2SO4, not just part of it. How much excess H2SO4 did we have? That's M H2SO4 = mmols/mL. M = 0.029297. mL in the final solution is 25.3 mL + 18.8 mL...

**Chemistry**

Yes, (l).

**Chemistry**

volume = (edge length)^3 = ? cm^3 mass = volume x density. You know volume and density. Solve for mass. Note that g/mL = g/cc = g/cm^3

**@Connexus cheat - banned**

Technically it is cheating; however, I'm not sure that's the way the students see it. I think in many cases it's a way to get someone to do their work for them. They are just too lazy to fill in the blanks/check the multiple guess/read the lesson/etc. and they see ...

**Science**

You can get better service if you don't change screen names.

**Chemistry**

delta T = Kf*m Kf = 1.86 m = mols/kg solvent and mols = grams/molar mass. Substitute and solve for m. Then substitute into dT = KfM and solve for dT. Post your work if you get stuck.

**Science**

Fe + S ==> FeS synthesis

**Chemistry**

The smallest set of whole numbers is C4H5N2O

**Chemistry**

See your other post.

**Chemistry**

2NaCl + 2H2O ==> Cl2 + H2 + 2NaOH coulombs = amperes x seconds = approx 2400 96,485 coulombs will form 40 g NaOH. g NaOH produced = 40 x 2400/96,485 = 0.995. How many mols is that? 0.995/40 = approx 0.25 Then you know the titration is NaOH + HCl ==> NaCl + H2O Approx (...

**chemistry**

I take it that 1.33E-10 is the average of Kb for acetic acid and you want to calculate Ka. Ka*Kb = Kw = 1E-14 You know Kw and Kb, solve for Ka.

**Chemistry**

Al is in group IIIA and is +3, I think, almost always. But you can tell by looking at the formula of Al2(SO4)3. The tells you oxidation number of Al is +3 and the 2 tell you the sulfate ion is -2. How to do S. Al is 2*+3 each = +6. O is 12*-2 = -24 Remember Al2(SO4)3 is zero ...

**Chemistry - Significant Figures**

rate = delta grams/delta time = 2.4/35 = ?

**chemistry**

http://www.jiskha.com/display.cgi?id=1495141830

**chemistry**

I don't know why it won't. You may have used the wrong color litmus paper to start.

**chemistry under gas law**

0.08206 x 101.25 kPa/atm = 8.314 J/mol

**chemistry ( little urgent pls)**

Yes, and that means dH must be large(negative) eough to off set dS. BTW, a little urgent is like being a little pregnant.

**chemistry**

PV = nRT. So if you know P, V, T and R, what's left?

**Chemistry**

Yes. Here are the definitions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Oxidizing agent and reducing agents are the reverse. Oxidizing agent is reduced; therefore, it gains electrons. Reducing agent is oxidized; therefore, it loses electrons.

**chemistry**

Water is polar. Ethanol is polar. Hexane is non-polar. Like dissolves like.

**chemistry**

See your other post. Like dissolves like.

**Chem**

You need to find the arrow key and use it. CaC2(s) + 2H2O(l)=> Ca(OH)2(aq) + C2H2(g) Use the coefficients to convert anything to any other reactant or product. 4.7 mol CaC2 x (1 mol C2H2/1 mol CaC2) = 4.7 x 1/1 = 4.7 mol C2H2. Note that the units you don't want cancel (...

**Chemistry**

1 is a little tricky. They way you do it is to balance the redox part, then the tricky part. Cu is 0 on the left and +2 on the right. Draw a line from one to the other and show loss of 2e. Then you look at N. It is +5 on the left and it shows +5 on the right (with the NO3^-) ...

**Science**

Nothing is shown. I don't know where points A and B are.

**chemistry**

NH4OH + HClO4 ==> NH4ClO4 + H2O millimols NH4OH = M x L = 6 mmols HClO4 = M x L = 6 So the acid and base exactly neutralize each other and you have the NH4ClO4 in aqueous solution at the end of the reaction. The concentration of the salt is mmols/mL = 6/60 = 0.1 M. The pH ...

**chemistry**

We prefer you not change screen names. This is worked the same way as the NH4ClO4 problem BUT the LiOH and HNO2 don't neutralize each other completely so you have one of them left over and that determines the pH.

**Chemistry**

See my response above.

**MJ Science 3**

https://www.thoughtco.com/properties-basic-metals-element-group-606654

**MJ Science 3**

https://en.wikipedia.org/wiki/Chemical_bond#Strong_chemical_bonds

**chemistry**

http://www.jiskha.com/display.cgi?id=1401238255

**MJ Science 3**

D is not correct for 2.

**Science**

It would help if we knew what you were talking about. Outputs and side effects of WHAT?

**Chemistry**

You add the states. You don't say acid or base solution. Ag ==> Ag^+ + e 3e + CrO4^2 + 8H+- ==> Cr^3+ + 4H2O