Friday
July 1, 2016

Posts by DrBob222

Total # Posts: 52,607

Chemistry
2H2 + O2 ==> 2H2O Use the coefficients to calculate this. 27.4 mols H2 x (1 mol O2/2 mols H2) = 27.4 x 1/2 = ?
March 29, 2016

Chemistry
Do you have choices? Ga^3+ would be smaller but so would many other ions.
March 29, 2016

chemistry
NaC2H3O2 = NaAc acetic acid = HC2H3O2 = HAc NaAc diluted = 0.293 x (190/750) = approx 0.074 but you need to confirm this and all other calculations that follow. I estimate and round. .......Ac^- + HOH ==> HAc + OH^- I....0.074............0......0 C.......-x............x...
March 29, 2016

AP Chemistry
I answered this at your other post above (with anonymous screen name).
March 29, 2016

chemistry
This is a limiting reagent (LR) problem; you know that because amounts are given for both reactants. 3Mg(s) + N2(g) --> Mg3N2(s) The long long way to do this. mols Mg = 8 from the problem. mols N2 = 2 from the problem. Using the coefficients in the balanced equation, ...
March 29, 2016

Chemistry- Dr.Bob222
No point in my weighing in!
March 28, 2016

Chemistry
CaCO3(s)-> CaO(s) + CO2(g) Convert 71.42 g CaCO3 to mols. mols = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO. Now convert mols CaO to grams. g CaO = mols CaO x molar mass CaO. This is the theoretical yield (TY). The ...
March 28, 2016

Chemistry
No. Your error is you convert mols NaOH to grams NaOH then used GRAMS for mols in M = mols/L but you can't do that. You should convert 2 g to mols KHP, then to mols NaOH, then mols NaOH/M = L NaOH
March 28, 2016

chemistry
8.5E2 cal x (1 Cal/1000 cal) = ? 0.45 kJ = 450 J 450 J x (1 cal/4.184 J) = ?
March 28, 2016

Chemistry
Ben, I did this far below. http://www.jiskha.com/display.cgi?id=1459199032
March 28, 2016

Chemistry
You explanation still doesn't make sense to me. reducing oxide? number of this oxidation reaction?
March 28, 2016

@ Korina--Chemistry
You need to make it clear as to what you want. Also it helps if you explain exactly what you don't understand.
March 28, 2016

Chemistry
see http://www.jiskha.com/display.cgi?id=1459199032
March 28, 2016

Chemistry
Ptotal = pH2O + pgas Ptotal = 100.2 kpa pH2O--look up vapor pressure H2O at 30 C (in kPa or convert to kPa or convert 100.2 kPa to mm Hg and use pH2O in mm Hg) Substitute and solve for pgas.
March 28, 2016

Chemistry
Look up Ksp for Mg(OH)2. .........Mg(OH)2 --> Mg^2+ + 2OH^- I........solid.......0........0 C........solid.......x........2x E........solid.......x........2x Substitute the E line into the Ksp expression for Mg(OH)2 and solve for x = solubility in mols/L. Convert to g. g = ...
March 28, 2016

Chemistry
Buffered at what pH. pH low the solubility is increased significantly. pH high, solubility is decreased significantly.
March 28, 2016

Chemistry
I have no idea what you're talking about. Valley?
March 28, 2016

Chmistry
What's your problem in doing this. The problem tells you the products they want. a) 2C2H2 + O2 ==> 4C + 2H2O b) 2C2H2 + 5O2 ==> 4CO2 + 2H2O c and d are done the same way.
March 28, 2016

Chemistry
First determine the M of the concentrated (the 37 %) HCl solution. That is 1000 x 1.19 x 0.37 x (1/36.45) = approx 12 M but you need to do it more exactly. Then use mL1 x M1 = mL1 x M2 mL1 x 12M = 1000 mL x 2.5M Solve for mL1 which is mL of the 37% stuff.
March 28, 2016

Chemistry
2Mg + O2 ==> 2MgO mols Mg = grams/atomic mass = 12.01/24.3 = ? mols O2 = grams/molar mass = 6.56/32 = ? Using the coefficients in the balanced equation, convert mols Mg to mols MgO produced. That's #1. #2. Convert mols O2 to mols MgO. #3. In limiting reagent problems ...
March 28, 2016

Chemistry
How many moles do you need? That's mols = M x L = ? Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
March 28, 2016

Chemistry
I'm not familiar with the experiment you are doing but I would go with the second equation with a +. Again, I'm not sure of the - sign in front of the second equation since I don't know how it's being measured.
March 28, 2016

Chemistry
(0.2*10) + (0.8*11) = mass B. To be accurate you should have the masses of 10B and 11B given and plug those values in instead of 10 and 11.
March 28, 2016

chem
Use molar mass and H bonding to determine. The higher molar mass (more van der Walls forces) the higher the b.p. The more H bonding (polarity) the higher the b.p. Butane, for example, would be the lowest b.p. You can look these up on google and see if you have them right.
March 28, 2016

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

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