Wednesday
May 25, 2016

Posts by DrBob222

Total # Posts: 52,429

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

Chemistry
Look at the oxidation state of each element on both sides. If they are the same redox does not occur. If they are different it is a redox reaction.
March 18, 2016

chemistry
You have the answer. What else is to explain? The AgSCN ppts, the SCN concn is decreased, the reaction shifts to the left to compensate and less FeSCN^2+ complex is formed.
March 18, 2016

Chemestry
Will, you need to learn how to spell chemistry. Volume of a solid makes no sense to me unless there is more to the problem.
March 18, 2016

chemistry
I don't know what experiment you did but if the pressure of the atomosphere was the same then smaller temperature means smaller volume.
March 18, 2016

chemestry
See the above and spell chemistry correctly.
March 18, 2016

chemestry
Oxidation is the loss of electrons. Which is losing electrons. A. Pb is +4 on the left and +2 on the right. That's a gain of electrons so A can't be right. B. O doesn't change oxidations state. B can't right. C. H doesn't change oxidation state so C can'...
March 17, 2016

chemestry
Louren, are you paying attention to the answers? Please use the correct spelling of chemistry. A. H2 has an oxidation state of zero on the left and +1 on the right so it is losing electrons. That makes it oxidized so A can't be right. B. If H2 is being oxidized then Cl2 ...
March 17, 2016

Chem101
Hydrocarbons + oxygen give CO2 and H2O. C7H16(l) + 11O2(g)→ 7CO2(g) + 8H2O(g)
March 17, 2016

Chemistry
dGrxn = dGo + RTlnQ dGo = -RTlnKa dGo = about +18.5 kJ RTlnQ = about -31.5 kJ dGrxn = about -13 kJ/mol
March 17, 2016

Chemistry
dGrxn = dGproducts - dGreactants Then dGrxn = -nFEo Solve for Eo.
March 16, 2016

chemistry
The hint tells you how to work the problem. mols NaOH = M x L = ? mols acid = mols NaOH since it is 1:1 ratio. Then M acid = mols acid/L acid = ?
March 16, 2016

Chemistry
This done the same way as you other acid/NaOH problem. The only difference is in this problem the acid is given a name and they want grams and not M. KHP + NaOH ==> NaKP mols NaOH = M x L = ? mols KHP = mols NaOH Then mols KHP = grams KHP/molar mass KHP. You know molar mass...
March 16, 2016

chemistry
I assume you want to prepare 500 cc of the CuSO4. How many mols do you want? That's M x L = 0.1 x 0.500 = ? Then convert to grams. g = mols x molar mass = ?
March 16, 2016

Chemistry DR BOB
I assume you mean the reaction in which electrons flow from the anode to the cathode spontaneously. Anode is on the left, it is oxidized; therefore, Eo oxidation = -0.37 Cathode is on the right so it receives the electrons and it is reduced so Eo redn = 0.67. EoCell = Eox + ...
March 16, 2016

Analytical Chem
Use PV = nRT and solve for pSO3 at the beginning and at equilibrium. I obtained about 1 for pSO3 initially but you need to get a more accurate answer. Then ..........2SO3 ==> 2SO2 + O2 I.........1.0.......0......0 C.........-2x.......2x.....x E........1.0-2x.....2x.....s ...
March 16, 2016

chemestry
Just as an add on, you really need to learn to spell chemistry correctly.
March 16, 2016

chemestry
Let's think about this. If a neutral solution is 1E-7 M for (H^+) and 1E-7 M for (OH^-), that means for the solution to be basic the (OH^-) must be larger than 1E-7 M. Is 4.2E-8 M larger than 1E-7 M?
March 16, 2016

chemestry
The correct answer is "none of these" which is not a choice. The answer your prof wants is B but that isn't correct. Many texts, if you can find the problem addressed, explains this very well but many texts don't even address this segment of chemistry. I have...
March 16, 2016

chemestry
Yes, A is the correct answer. In the acid, HA, it ionizes as HA ==> H^+ + A^- Ka = (H^+)(A^-)/(HA) The acidity of a solution depends upon the (H^+) so the higher that is the stronger the acid. Look at Ka values. If Ka is large then the numerator (the H^+) is larger. For the...
March 16, 2016

Chem
XH2SO4 = 0.2 so XH2O = 0.8 Take 1 mol of the solution. That will contain 0.8 mols H2O. How many grams is that? 0.8 mol x 18 gH2O/mol = 14.4 g or 0.0144 kg. molality = mols H2SO4/kg solvent 0.2/0.0144 = ?m
March 16, 2016

Chemistry
Yes, that final post looks good. Good work.
March 15, 2016

Chemistry
You need to learn to do these by yourself and me doing them will not help you. Here is a site that will help or if you have trouble show what you can do and explain fully what you don't understand about the next step. Cr goes from +6 on the left(for each) to 3+ on the ...
March 15, 2016

Chemistry
The numbers may be slightly different but the process is the same. http://www.jiskha.com/display.cgi?id=1458010856
March 15, 2016

Chemestry
I would convert 100 mM to M and mL to L. Then, how many mols do you need? That's mol = M x L = ? Next step is grams = mols NaCl x molar mass NaCl = ?
March 15, 2016

Chemestry
See your other post. Same process.
March 15, 2016

Chemistry
pH = 9 means pOH = 5 so (OH^-) = 1E-5 Substitute and solve for Qsp. Qsp = (Cd^2+)(OH^-)^2 and compare that with Ksp. If Ksp is smaller there is a ppt. If Ksp is larger there is no ppt.
March 15, 2016

Chemistry
I assume the 0.145m is a typo and you meant 0.145 M. Also I assume you meant NaOH and not naoh which means nothing to me. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O mols H2SO4 = M x L = ? Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH. That's mols...
March 15, 2016

chemestry
no. A has the least.
March 15, 2016

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