Thursday
September 29, 2016

Posts by DrBob222

Total # Posts: 53,178

chemistry
acetic acid is HC2H3O2 molar mass 60 Oxalic acid is H2C2O4 molar mass 90. Oxalic acid is both more London forces as well as more H bonding.
March 28, 2016

Chemistry
N = #H or OH*M or M = N/#H or OH For H3PO4. 6N/3 = 2M For Ca(OH)2. 4N/2 = 2 etc. M =
March 28, 2016

Chemistry
(p1v1/t1) = (p2v2/t2) Remember T must be in kelvin. Remember to use the same units for p.
March 28, 2016

Chemistry
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. 2Fe + 3O2 ==> Fe2O3 mols Fe = grams/atomic mass = ? mols O2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Fe to mols Fe2O3. Do the ...
March 28, 2016

Chemistry
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants. Your first step is ok. Using the coefficients in the balanced equation convert mols CaCO3 to mols CO2. Do the same and convert mols HI to mols CO2. It is likely that these two ...
March 28, 2016

Equilibrium Consentration
H2 + Cl2 ==> 2HCl Kc = (HCl)^2/(H2)(Cl2) You know Kc, (H2) and (Cl2). The only unknown is (HCl). Substitute and solve for that.
March 28, 2016

Chemistry
CH3NH3Br ==> CH3NH3^+ + Br^- Then the CH3NH3^+ is hydrolyzed ..CH3NH3^+ + H2O ==> CH3NH2 + H3O^+ I..0.01...............0........0 C...-x................x........x E.0.01-x..............x........x Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.01-x). Solve for x = (...
March 28, 2016

chemistry12_2016-4
See your Mg/HCl post below. Follow the directions. Post your work if you get stuck.
March 27, 2016

chemistry12_2016-3
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. The following steps will work almost any LR problem. 1. Write and balance the equation. Mg + 2HCl ==> MgCl2 + H2 2. Convert reactants to mols a. mols HCl = M x L = 0.5 x 0....
March 27, 2016

chemistry12_2016-2
M = mols/dm^3 = mols/0.1 mols = grams/molar mass = 14.0/approx 56 = ? Substitute into M = mols/L and solve for M.
March 27, 2016

Chemistry
A poorly worded question. What you want to know is how much H2SO4 was used, not how much xs was used. Technically, none of the excess was used. Did your problem provide an equation. Check and see that it looks like this. Ca5(PO4)3F + 5H2SO4 ==> 5CaSO4 + 3H3PO4 + HF If my ...
March 27, 2016

chemistry12_2016-1
RTP is Room Temperature & Pressure. Room T is considered 298 K.
March 27, 2016

chemistry12_2016-1
Another limiting reagent problem Follow the directions in your posts above. Post your work if you get stuck and I can help you through it.
March 27, 2016

Chemistry
Thank you for the equation. See your post above and change the equation I proposed to the proper one. Post your work there if you run into trouble and I can help you through it. For this problem, mols Ca5(PO4)3F = grams/molar mass = 5/504.3 = approx 0.006 but you need a more ...
March 27, 2016

Chemistry
You want me to write a term paper for you?... No thanks.
March 27, 2016

Chemistry
Look up Graham's Law. The rate is proportional to the square root of the molar mass.
March 27, 2016

Chemistry
ptotal = pH2 + pH2O ptotal = 757 pH2O = 17.5 pH2 = ? Substitute and solve.
March 27, 2016

Chemistry
I don't understand anything you wrote. Are you working this problem or another one? Plug values into that equation I gave you and solve. What you have posted is gibberish.
March 27, 2016

Chemistry
(p1v1/t1) = (p2v2/t2) Remember T must be in kelvin.
March 27, 2016

chemistry
Use the H-H equation. Technically one should use concentrations but I like to work in millimols = mL x M. And since both numerator and denominator for M = millimols/mL and mL is the same (because it's the same solution) then using millimoles gives the same answer. mmols ...
March 27, 2016

chemistry
Do you have any other information? You need, at least, the heat of neutralization for H^+ + OH^- ==> H2O. I don't remember but think it's about 55 kJ/mol. HCl + NaOH ==> NaCl + H2O You have mols HCl = M x L = 0.1 x 0.1 = 0.01 mols. So approx 55 kJ/mol x 0.01 mol...
March 27, 2016

chem
Use PV = nRT and solve for n = number of mols air in the lungs. Then n x 0.21 = mols oxygen.
March 27, 2016

Chemistry
I should point out that m and M are not the same. I assume that is a typo. It isn't easy to convert m to M with knowing the density.
March 27, 2016

CHEM QUESTION!
A very easy way to do these ppm problem in water solution is to remember that 1 ppm = 1 mg/L or 1 ug/mL. So 0.6 ug/mL = 0.6 ppm.
March 26, 2016

Chemistry
I think the correct solution is to recognize this as a buffer problem use the Hendersib-Hasselbalch equation. pH = pK2 + log (base)/(acid) pH = 1.88 + log (0.574/0.2) = ? pH approx 2.3 OR you can work it as a common ion; i.e., HSO4^- ==> H^+ + SO4^- k2 = ? K2SO4 ==> 2K...
March 26, 2016

Chemistry
The first half equation is not balanced. Cl on the left is 7+ and on the right is 1- so delta e is 8e on the left. You will notice the charges don't balance in your half equation. 1+ on the left and 1- on the right.
March 26, 2016

Chemistry
Em, I answered this only 10 minutes after you posted it first. Give us a break.
March 26, 2016

Chemistry
.....H2 + Cl2 ==> 2HCl E...0.26..0.087....x Kc = (HCl)^2/(H2)(Cl2) You know Kc, (H2), and (Cl2), substitute and solve for HCl.
March 26, 2016

~ CHEMISTRY ~
The first two lines ask a question. If P goes up, V goes down. I don't know what you want for the rest of the post.
March 25, 2016

Chemistry
2H2 + O2 ==> 2H2O 1.249 g + 9.99 g = ? The law of conservation of mass says that you get out what you put in.
March 25, 2016

AP Chemistry
I think I answered this for you yesterday. Use the Henderson-Hasselbalch equation.
March 25, 2016

chemistry
Here is the way to do these. The reaction is this. ....M^2+ + 4CN^- ==> [M(CN)4]^2- I..0.170...1.04.......0 C.-0.170..-4*0.170 ...0.170 E...0.....0.36.......0.170 You can see that with a K of a bouat 10^16 that the reaction is essentially complete to the right. Now you ...
March 25, 2016

chemistry
I looked up the density of a 37% solution at 20 degrees C = 1.18 Then 1000 x 1.18 x 0.37 x (1/36.45) = 12.3 M for the HCl. Then mL1 x M1 = Ml2 x M2 mL1 x 12.3 = 1000 mL x 0.2 assuming you want to prepare 1000 mL of the solution. So you take mL1 you calculate, add to a 1000 mL ...
March 25, 2016

Chemistry
mL1 x M1 = mL2 x M2 mL1 x 12 = 2000 mL x 3.0 M Solve for mL1
March 25, 2016

Chemistry
There is 1 mol N in 1 mol HNO3. 1 mol N = 14 g IF you mean nitrogen gas as in N2, then there are 7 g N2 since there is 1/2 mol N2 in 1 mol N.
March 25, 2016

chemistry
To clarify, do you want 0.2m or 0.2M?
March 25, 2016

Chemistry
Using the 6.0 M stock. mL1 x M1 = mL2 x M2 mL1 x 6.0 = 250 mL x 0.01 Solve for mL 1, add that to a 250 mL volumetric flask and make to the mark with DI water. 2.5 M stock is done the same way.
March 25, 2016

chemestry
Use the coefficients to do this. 0.3 mol N2 x (2 mols NH3/1 mol N2) = 0.3 x 2/1 = ?
March 24, 2016

chemestry
See your other post and note the correct spelling of chemistry.
March 24, 2016

Chemistry
2 mols NH3 formed. 1 mol N2 and 3 mols H2 are represented. That is 6.02E23 molecules of N2 and 3*6.02E23 molecules of H2.
March 24, 2016

Chemistry
I don't see a question here.
March 24, 2016

Chemistry
I don't see a question here.
March 24, 2016

chemistry
........NiCO3 ==> Ni^2+ + CO3^2- I.......solid.....0........0 C.......solid.....x........x E.......solid.....x........x Substitute the E line into the Ksp expression and solve for x = (NiCO3) in mols/L. Then grams = mols x molar mass = ? Compare the solubility with the 5 mg...
March 24, 2016

chemistry
.......CaCrO4 ==> Ca^2+ + CrO4^2- I......solid.......0.......0 C......solid.......x.......x E......solid.......x.......x ........Ca(NO3)2 ==> Ca^2+ + NO3^2- I........0.25M........0.......0 C.......-0.25.......0.25......0.25 E..........0........0.25......0.25 Note that ...
March 24, 2016

Chemistry
It is correct if you punch in the right numbers. Your set us is right; the answer is wrong.
March 24, 2016

Chemistry
Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid) 9.0 = 9.24 + log 0.1M/(acid) Solve for (acid) in M. Then M = mols/L. You know M and L, solve for mols. Then mol = grams/molar mass. You know mol and molar mass, solve for grams. Piece of cake.
March 24, 2016

Chemistry (30s)
The empirical formula mass is 15; i.e., 12 for C + 3*1 for H = 15. The molecular formula mass is 30 g so the question is asking how many units of CH3 make up the molecule. That's 30/15 = 2 or the molecular formula is(CH3)2 = C2H6. You can check it this way. 2*12 + 6*1 = 30
March 23, 2016

AP CHEM
Yes. 7.45 = pKa2 + log (b/a) b/a = ?
March 23, 2016

AP CHEM
You're trying to make this a hard problem when it isn't. This is a buffered solution, use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid) They give the pH, base is 0.85 and acid is 0.65. Substitute and solve for pKa.
March 23, 2016

AP Chemistry
See your other post.
March 23, 2016

AP Chemistry
strong base, NaOH, and week acid, HF. >7. approx 8.5 c. Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid)
March 23, 2016

Chemistry
Chris, Terry, and the others. It really helps us help you if you use the same screen name each session. I don't know what data you have available in your tables but I would try dGorxn = (n*dGoproducts - n*dGoreactants) and if dGrxn < 0, then K>1. For the last part, ...
March 23, 2016

Chemistry
Use the van't Hoff equation.
March 23, 2016

Chemistry
mols CO = 1 mols Cl2 = 1 total mols = 2 XCO = 1/2 = 0.5 XCl2 = 1/2 = 0.5 pCO = XCO * Ptotal = 0.5 x 1 = 0.5 atm pCl2 = XCl2 * Ptotal = 0.5 x 1 = 0.5 atm .......CO(g) + Cl2(g)-->COCl2(g) I......0.5.....0.5........0 C.......-x......-x........x E....0.5-x.....0.5-x......x ...
March 23, 2016

Chemistry
I assume this refers to your other question below. mols H2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols H2 to mols Zn. Now convert mols Zn to grams, then to kg. grams = mols Zn x atomic mass Zn = ?
March 23, 2016

Chemistry
Zn + H2SO4 ==> H2 + ZnSO4 The equation is balanced as is. If you want the phases, it looks like this. Zn(s) + H2SO4(l) ==> H2(g) + ZnSO4(aq) with some uncertainty about the phase of H2SO4. Most H2SO4 out of the bottle has a little water in it and some may prefer to call ...
March 23, 2016

chemistry
2Al + 6HBr ==> 3H2 + 2AlBr3 mols Al = grams/atomic mass = ? Using the coefficients in the balanced equation, convert mols Al to mols H2 gas. Now convert mols H2 gas to liters knowing that 1 mol occupies 22.4 L at STP.
March 23, 2016

Chemistry
I'm a little confused about the question because you show the specific heats but don't list a starting T. So I will ignore those specific heats. Then dG = dH - TdS dG at freezing is an equilibrium and = 0 so 0 = dH - TdS dH is 2688 T is -25. Convert to K. Solve for dS.
March 23, 2016

Chemistry
N2 + 3H2 ==> 2NH3 mols N2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols N2 to mols NH3. That will be twice mols N2 = mols NH3. Then grams NH3 = mols NH3 x molar mass NH3.
March 23, 2016

CHS Chemistry
.......MCl2 ==> M + Cl2 .....73.79g..45.25..28.54 mols Cl2 = 28.54/molar mass = approx 0.4 but you may need to do it more accurately. Since 1 mol Cl2 is given off for 1 mol MCl2 and 1 mol M, then 45.25 g M must be approx 0.4 mol. mol = grams/atomic mass. 0.4 45.25/atomic ...
March 23, 2016

Chemistry
NH4Cl ==> NH3 + HCl mols NH3 produced = grams/molar mass = 134/17 = approx 8 but you need a better answer. Using the coefficients in the balanced equation A(everything is 1:1) convert mols NH3 to mols NH4Cl. Now convert mols NH4Cl to grams. g NH4Cl = mols NH4Cl x molar mass...
March 23, 2016

chemistry
What about it. Go through the same reasoning and I'll check it for you. Remember the definitions. Oxidation is the loss of e. Reduction is the gain of e. The material oxidized is the reducing agent. The material reduced is the oxidizing agent.
March 23, 2016

chemistry
I know what you're asking but I think you gave the wrong example. Is that SO3^2- ==> SO4^2-. If so, then S on the left is +4 and S on the right is +6 so the change is -2 electrons. Since oxidation is the loss of electrons then SO3^2- must be oxidized. That means SO3^2- ...
March 23, 2016

Chemistry
volume of 10,000 cc is correct. 10,000 = 14.5 x 14.5 x thickness Solve for thickness. 690 cm sounds way too large to me.
March 23, 2016

Chemistry
Co(ClO4)2 ==> Co^2+ + 2ClO4^- You know 1 mole of anything contains 6.02E23 molecules. So how many moles do you have of this stuff. mols = grams/molar mas so mols = 6/257.83 = ? So ? x 6.02E23 = number of molecules of Co(ClO4)2. You want to know ClO4^-. There are two of ...
March 23, 2016

Chemistry
Almost. You have the right idea but you've reversed the facts. The empirical formula is the simplest; the molecular formula is how many of the empirical units you have together. So the empirical mass is 2*B + 5*H or 2*10.81 + 5*1 = 26.62 The molecular mass is 53.3 from the...
March 23, 2016

Chemistry
This is a limiting reagent (LR) problem and a percent yield rolled into one. You know it is a LR problem because amounts are given for BOTH reactants. mols H2 = grams/molar mass = ? mols O2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols H2...
March 23, 2016

Chemistry
35.9 g NaCl x (1000 mL/100 mL) = 359 g NaCl that will dissolve in 1 L of H2O. Anything over that will not dissolve and the solution will be saturated. How much is left? That's 1000 g NaCl - 359 g NaCl = ?
March 23, 2016

chemistry
This is a limiting reagent (LR) problem since an amount is given for BOTH reactants. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g) Starting with 5.00 mols NO2 and all of the water you need will produce 5.00 x (2 mol HNO3/3 mols NO2) = 5*2/3 = about 3.3 Starting with 2.00 mols H2O ...
March 23, 2016

Chemistry
Close but no cigar. It is not a gram that contain 6.022E23 but a MOLE that contains 6.022E23. So convert that 1.32 g to mols. mols = grams/molar mass = ?. Then that many mols x 6.022E23 molecules/mol = # molecules of C10H8..
March 23, 2016

Chemistry
Many of your numbers are right but they are in the wrong place. Here is what you should have done. mols H2 = 2.23/2.016 = 1.106 mols I2 = 0.218. So these two are ok. mols HI produced from H2 = 2*1.16 = 2.21 mols HI produced from I2 = 2*0.218 = 0.436 and you are right that I2 ...
March 22, 2016

Chemistry
Yes, it is a limiting regent (LR) problem with and extra percent yield thrown in. H2 + I2 ==> 2HI mols H2 = grams/molar mass = ? mols I2 = grams/molar mass = ? Now, using the coefficients in the balanced equation, convert mols H2 to mols HI produced. Do the same and convert...
March 22, 2016

Chemistry lab
Technically we can't do this because you don't specify an indicator; therefore, we don't know how many of the H ions were titrated. I will assume we titrated all 3. H3A + 3NaOH ==> 3H2O + Na3A mols NaOH = M x L = ? Then ? mols NaOH x (1 mol H3A/3 mols NaOH) = ...
March 22, 2016

Chemistry
Yes it does. Rounding to the nearest whole number you have C1, F1, Cl3 and I made a typo. So the empirical formula is CFCl3. I'm glad you caught that.
March 22, 2016

Chemistry
NOPE. You divide, not multiply. mols C = 8.74/12.01 = ? mols F = 13.8/19 = ? mols Cl = 77.4/35,45 = ? Then find the ratio as you did. I get CFCl2.
March 22, 2016

Chemistry
NaCl, KBr, NaClO4 CANNOT act as a buffer. The others can.
March 22, 2016

Chemistry
I thought I answered this for you last week. HBr, NaOH, HClO4.
March 22, 2016

Chemistry
Dinitrogen monoxide is N2O. If you have five of those then you must have 5*2 = 10 N atoms. b. 1 mol of N2O contains 6.022E23 molecules so 5 mols will contain 5*6.022E23 N2O molecules of N2O.
March 22, 2016

Chemistry
b is correct. a. A single molecule has a mass of 128.15 amu c. Since there are 6.022E23 molecules in a mole (128.16g), then 1 atom has a mass of 128.16/6.022E23 = ?
March 22, 2016

Chemistry
You are absolutely correct. Good work.
March 22, 2016

Chemistry
Why don't you show how you did it ans let us check your work.
March 22, 2016

Chemistry
Almost. a. Yes, 11 mols of atoms (3 Ca, 2P, 6O = 11 total) b. You have 2 mols PO3^2- (that subscript of 2 after the closed parentheses says 2. c. I think c is asking for individual PO3^3-, which I missed the first time around. That means you have 6.02E23 ions in 1 mol so twice...
March 22, 2016

Chemistry
You use the subscripts to tell you.I will assume you made a typo and you meant to type in Ca3(PO4)2 but the instructions are the same for Ca3(PO3)2. If you have 1 mol Ca3(PO4)2 you have 6.02E23 molecules of Ca3(PO4)2. You have 3 mols of Ca ions, 2 of P, 12 of O and 3 of PO4^3-.
March 22, 2016

Chemistry
That's a good start. mols Pb(NO3)2 = grams/molar mass = 125/331.2 = 0.377 mols KI = 125/166 = 0.753 Now the next step is to convert mols of each to mols of the product. You say the directions I gave are confusing. What do you not understand about them? You start with 0.377...
March 22, 2016

Chemistry
That isn't right and you didn't follow what I wrote at the beginning. mols Pb(NO3)2 = grams/molar mass = ? mols KI = grams/molar mass = ?
March 22, 2016

Chemistry
OK. How many mols Pb(NO3)2 do you have? How many mols kI do you have? What about the converting using coefficients do you not understand?
March 22, 2016

Chemistry
This is a limiting reagent (LR) problem. You know that because BOTH reactants are listed. Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3 mols Pb(NO3)2 = grams/molar mass = ? mols KI = grams/molar mass = ? Using the coefficients, convert mols KI to mols of PbI2. Do the same and convert ...
March 22, 2016

Chemistry
I'm sorry but I just don't have all of the solubility tables memorized. There must be millions. You must have a table or a graph and you can read it from that.
March 22, 2016

Chemistry
HCl + NaOH ==> NaCl + H2O mols NaOH = M x L = 0.1 x 0.02 = 0.002. mols HCl = 0.002 from the coefficients in the balanced equation. M HCl = mols/L = 0.002/0.02 = 0.1M HCl. Convert to pH. pH = -log(H^+). H2SO4 + 2NaOH ==> 2H2O + Na2SO4 mols NaOH = M x L Using the ...
March 22, 2016

Grade 11 Chemistry
Menthal has nothing to do with methanol.
March 22, 2016

Grade 11 Chemistry
I'm not sure I've seen a diagram like this but I would draw a polar molecule of water and a polar molecule of CH3OH and show the H bond between the O of one and the H of the other. Several of those H bonds between the two molecules should show what is happening
March 22, 2016

Chemistry
Fe^3+ + SCN^- ==> FeSCN^2+ Keq = [FeSCN^2+]/[Fe^3+][SCN^-] So if SCN is not used up it will be too high and FeSCN^2+ will be too low. That will make Keq too low.
March 22, 2016

Chemistry
Look up the reduction potential for PbO2 ==> Pb(OH)4^2- and reverse the sign. Look up the reduction potential for the ClO^- ==> Cl^-. Add the oxidation E and redn E to find Eocell. Then dGrxn = -nFE
March 22, 2016

Chem
There are 6.02E23 atoms in a mol so in 0.002 mols there are 6.02E23*0.002 = ? Since there are 4 atoms/H2SO4 molecule, multiply the previous number by 4.
March 22, 2016

chemistry
1 mol contains 6.02E23 so there are 1.51E24/6.02E23 = ? mols H2O.
March 22, 2016

Chemistry
Let's call lactic acid HL. .......HL --> H^+ + L^- I.....0.05....0.....0 C.....-x......x.....x E.....0.05-x..x.....x Substitute the E line into Ka expression and solve for x = (H^+), then convert to pH. dG = -RTlnKa
March 22, 2016

chemistry
Same type problem as your CH4 problem below.
March 22, 2016

chemistry
The kJ for 16 grams CH4 is -890.3. How many mols CH4 do you have? That's PV = nRT. How many grams is that of CH4? That's n = grams/molar mass. You know molar mass and mols, solve for grams CH4, Then -890.3 kJ x (grams CH4/16) = ? kJ for that reaction for that many ...
March 22, 2016

@ Steve--Chemistry
No, that gives you mols/L which is mols/dm^3. 0.0667 mols/25 cc. Convert to 1000 cc (1 dm^3) as above.
March 21, 2016

Chemistry
mols X in 25 cc = 4.0/60 = 0.0667 Then 0.0667 mols x (1000/25) = ?
March 21, 2016

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