Wednesday
July 27, 2016

Posts by DrBob222

Total # Posts: 52,756

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

chemistry
If it uses all 6 oxygen atoms, I would go with hexadentate. See https://en.wikipedia.org/wiki/18-Crown-6
March 21, 2016

Chemistry
I don't think the question is asking you to determine dGrxn. I think the question is asking, "Is the rxn spontaneous or not?" and the answer is yes it is. If you look at dStotal = dSsurr + dSrxn dSrxn is + since liquid H2O has higher entropy than solid H2O and ...
March 21, 2016

Chemistry
What's your problem with these? You need to learn the functional groups.
March 21, 2016

Chemistry
The answer I assume you want is electrons, protons, neutrons. But I have news for you. That's what I taught for years (and was taught for years) but the only REALLY fundamental particle in those three is the electron (as far as we know today). What I'm saying is that ...
March 21, 2016

Chemistry
How many mols do you want? That's M x L = mols. Then mols = g/molar mass. You know mols and molar mass, solve for grams.
March 21, 2016

chemistry
You don't have the entire problem copies. What you need is the volume of NaOH added in your titration but here is what you do. mols NaOH = M x L. That's 0.75M x L you used in the titrant. mols HCl = mols NaOH from above. Then M HCl = mols HCl/L HCl. The problem says L ...
March 21, 2016

Chemistry
I don't know what experiment you did.
March 21, 2016

chemistry
We try not to do your work for you. Use PV = nRT and solve for n = number of mols. Then n = grams/molar mass. You know grams and n, solve for molar mass.
March 21, 2016

Chemisrty
2KI(aq) + 2HOH(l) ==> H2(g) + 2KOH(aq) + I2
March 20, 2016

chemistry 1020
mL1 x M1 = mL2 x M2 67.0 x 0.4M = mL2 x 0.1M Solve for mL2 which will be the TOTAL volume. Since you had 67.0 mL initially, subtract from the total to find how much must be added.
March 20, 2016

Chemistry
Something is missing here; namely, how much Pb ion is present?
March 20, 2016

Chem
You need the density of the acetic anhydride.
March 20, 2016

Chemistry
Br^-, ClO4^-
March 20, 2016

Chemistry
Your equation is not balanced. I'll bet if you balanced it the stoichiometry would work. If it still doesn't work post you work and I'll find the error.
March 20, 2016

chemistry
a. 196 kJ x 2.65/2.00 = ? kJ produced. b. 2 mols H2O2 = 2*34 = 68 g. 196 x (234/68) = ? kJ produced.
March 20, 2016

Chemistry
I should point out here that since CaS is soluble in water that extenuation circumstances will be necessary (temperature, saturation points, etc) in order to produce a solid. % yield = 100*(actual/theoretical) 85% = 100*(3.0/theoretical) Theoretical = 3.0/0.85 = about 3.6 ...
March 20, 2016

Chemistry Honors
(0.580-0.238)/8 = ?
March 20, 2016

chemistry
(H^+)(OH^-) = Kw = 1E-14 You know OH, solve for H^+. If H = OH neutral If H > OH acid If H < OH basic
March 20, 2016

chemistry
That's right. 8.5 x 4 x 142 = ?
March 20, 2016

chemistry
No. How many mols do you want? That's niks = M x L = ? Then grams = mols x molar mass = ?
March 20, 2016

Chemistry
34 x (500/100) = ?
March 19, 2016

chemistry/statistics
And what is your problem with this. Can't you just substitute these numbers into the formula; i.e., plug and chug.
March 19, 2016

chemistry
N2 + 3H2 ==> 2NH3 If you want the volume of NH3 at the same conditions, it is just 23.7 L x (2 mols NH3/3 mols H2) = ? If you want it at STP, use PV = nRT and convert.
March 19, 2016

Chemistry
g NaOH = mols NaOH x atomic mass Na = ?
March 19, 2016

chemistry
C5H10 is correct for the alkene. The alkane will be C5H12 or pentane, isopentane, or neopentane. Wouldn't the alkanes be there in equimolar proportions; i.e., the mole fraction would be 0.5.
March 19, 2016

chemistry
You need to find the caps key and use it. 5E3 means 5 x 10^3 = 5000 = 5 times 10 to the third power. 1 mol contains 6.02E23 ions. 3E22/6.02E23 = approx 0.05 mol Cl^- ions. mols CaCl2 = 1/2 that since 1 mol CaCl2 contains 2 mols CaCl2. Then M = mols/L = approx 0.05 mols/0.200 L...
March 18, 2016

Chemistry
Bob Pursley is correct. I want to add some additional information here. A strong acid such as HCl or a strong base such as NaOH act as buffers by themselves. Not very good buffers but that solution IS buffered with just HCl. Look at the titration curve of a strong acid at this...
March 18, 2016

  1. Pages:
  2. <<Prev
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. Next>>