Monday
December 22, 2014

# Posts by DrBob222

Total # Posts: 46,158

Chemistry
The solubility of CaF2 is increased due to the presence of H^+ because H^+ confines with the F^- to form the weakly ionized HF. The easy to do this is as follows. First, determine the value of H^+) from the buffer part of the problem. pH = pKa + log (base)/(acid). You know pKa...
December 2, 2014

Chemistry
So write the Keq expression, substitute the concentrations given and solve for the concn of H2O. Post your work if you get stuck.
December 2, 2014

Chemistry
This is a problem that almost has to be worked "backwards" to make it work. .....Fe^3+ + 3C2O4^2- ==> [Fe(C2O4)3]^3- I....0.02....1.00...........0 C....-0.02..-0.06..........0.02 E......0.....0.94..........0.02 What I have done above is to look at the size of Kc (...
December 2, 2014

Chemistry
.....cis-butene==> trans-butene I......0.250........0.145 C.......-x...........+x E......0.250-x.....0.145+x Substitute the E line into Kp expression and solve for x, 0.250-x and 0.145+x. How did I know to make the C line -x......+x an not +x......-x; i.e., how do I know it...
December 2, 2014

Chemistry
If you care to post your work I shall be happy to go over it to find the error. But first, did you use Ea in Joules/mol and not kJ/mol (a common mistake) and did you cnvert T1 and T2 to kelvin (not so common error). The usual other problem is to reverse k1 and k2. Just make ...
December 2, 2014

Chemistry
Can't you use the Arrhenius equation for this. You have Ea, k1 along with T1 and T2; the only unknown is k2.
December 2, 2014

Since we have so many people dancing here let me dance too. Another reason, and just as important, is to utilize the common ion effect of the excess ion. In doing so you shift the equilibrium, just as Le Chatelier said will happen, and you DECREASE the solubility of the stuff ...
December 2, 2014

I think your answer is the one that the "prof" wants but I might clarify a little. Yes, NaNO3 is not easily determined both because most Na salts are soluble and most nitrate salts are soluble so there is no easy way to convert them to insoluble forms. Of course you ...
December 2, 2014

Honors Chem
First you must determine how much heat must be removed to make that temperature transition. q = [mass H2O x specific heat H2O x (Tfinal0Tinitial)]. q should be in J. Then calculate mols NH4Cl needed to get that q. 14,800 J/mol x # mols = q(in J) Then convert mols NH4Cl to grams.
December 2, 2014

chemistry
Potassium alum is KAl(SO4)2.12H2O 0.422g Al x (molar mass KAl(SO4)2.12H2O/atomic mass Al) = ? theoretical yield.
December 2, 2014

chemistry
Use PV = nRT
December 2, 2014

Chemistry
10 min/2 min half life = 5 So it has gone through 5 half lives. 2^5 = 32 10g initially/32 = ? what's left.
December 2, 2014

chemistry
This is done the same way as the Ba-122 problem above.
December 2, 2014

Chemistry
I think your problem is taken care of with the units. Since k has units of mol/L*atm that is M/atm and C comes out in M.
December 2, 2014

Chemistry
Do you recognize that this is a limiting reagent problem? That may be the problem. If you had posted what you did I could have pointed out the exact place you went wrong. Determine the limiting reagent. Using that substance, convert mols of the LR to mols of the product Then ...
December 2, 2014

Chemistry
0.5/0.70 = approx 0.714 g of the 70% stuff you need. Then convert grams to mL using density. Remember the %yield problems you worked early in chemistry? 70% stuff x what number = 0.5 g real stuff what number = 0.5/0.70 = about 0.7 g. Doesn't that make sense. You must take ...
December 2, 2014

Chemistry Problem
If you still want to see this done please post again at the top of the page and when you do note that the first sentence talks about diiodine pentafluoride but the reaction you give with the problem uses I2O5 (which I always called iodine pentoxide but I guess the proper name ...
December 2, 2014

Chemistry
How many mols S are in 5.60 g S? If call sulfur as S (and not S8), the mols S in 5.60 g S will be 5.60/atomic mass S = 5.60/32 = ? So you have that many mols of atoms. If you are to use S8 as sulfur, please post again at the top of the page and bring that to our attention.
December 2, 2014

Chemistry
Cd ==> Cd^2+ + 2e BrO3^- + 6e + 6H^+ ==> Br^- + 3H2O Br goes from +5 to -1 which is 6 so 6 e must be transferred. For Eo cell add the Cd part to the BrO3^- part (oxidation half cell to reduction half cell). For Ecell = Eocell -(0.0592/n)*log(Q) What's Q? That's (...
December 2, 2014

chemistry
Since L(lower case) can be any integer from 0 to n-1, then if L is 4 n must be 5. mL can be -L to + L which means -4, -3, -2, -1, 0, +1.......+4. Stick two electrons in each mL and count or get sophisticated and think it through. For s electrons, L(lower case) = 0, for p ...
December 1, 2014

Chemistry
I think the easy way to solve this problem is to recognize that the HCl titrates the Na2CO3 and leaves the NaCl as is. Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2 Total HCl added = M x L = approx 13.365 millimols but you need to confirm all of these numbers since I'm estimated ...
December 1, 2014

Chem 130
2CO + O2 ==> 2CO2 15.0 L CO at STP = 15.0/22.4 = ?mols Convert mols CO to mol O2, then use PV = nRT at the conditions listed to solve for volume (in liters) of O2.
December 1, 2014

Chemistry
You need to re-read the chapter because all I can do is repeat much of what your text has. But here is a summary. You have 1 proton in the nucleus. The single electron usually occupies the first shell which we will call n = 1. If energy is added, and there are various ways to ...
December 1, 2014

chemistry
Laila is El wrongo! Used the factor upside down to get 15.3. 2KClO3 ==> 2KCl + 3O2 mols KClO3 = mols O2 x (2 mols KClO3/3 mols O2) = 10.2 x (2/3) = ?
December 1, 2014

CHEMISTRY
(OH^-) = 10^-9 is correct.
December 1, 2014

Chemistry
The one LOSING H^+ is the acid that's the one DONATING the proton); the one GAINING H^+ is the base (that's the one ACCEPTING the proton.)
December 1, 2014

Chemistry
I'm surprise AND I disagree with you. Your posts show more work and better thinking than many of the students that post and in most cases you've worked through most of the problem but just stuck on one point near the end and need to jump that one hump. So don't get...
December 1, 2014

Chemistry
I answered the S + NO problem back on page 3 or 4.
December 1, 2014

Chemistry
Here is the way to think through this. Follow closely and put in the numbers where they belong. BaSO4 ==> Ba^2+ + SO4^2- Ksp = (Ba^2+)(SO4^2-) So when you start adding sulfate into the mixture, where will BaSO4 start to ppt? That's when Ksp for BaSO4 is first exceeded ...
December 1, 2014

AP Chemistry
Yes, CH4 + O2 ==> CO2 + 2H2O This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. mols CH4 = grams/molar mass mols O2 = grams/ molar mass Using the coefficients in the balanced equation convert mols CH4 to mols CO2. Do the ...
December 1, 2014

chemistry
no. pOH is 7 (but not molar). pH + pOH = 14 pH = 7; therefore, pOH = 7 Now do pOH = -log(OH^-) and solve for OH^-
December 1, 2014

chemistry
pH + pOH = pKw = 14. You know pKw and pH, solve for pOH. Then pOH = -log(OH^-) Solve for OH^-
December 1, 2014

Chemistry
It's 1.2937 but note that you're allowed only two significant figures (limited by the 2.0 mL) so that would round to 1.3 M.
December 1, 2014

Chemistry
H2SO4 + 2NaOH ==> 2H2O + Na2SO4 mols NaOH = M x L = ? Note that mols H2SO4 = 1/2 mols NaOH Then M H2SO4 = mols H2SO4/L H2SO4.
December 1, 2014

chemistry
Convert 13.2 g AlCl3 to mols. mols = grams/molar mass = ? Then M = mols/L solution. You know mols and M, solve for L and convert to mL.
December 1, 2014

Chemistry
Is that the equation given for the reaction? USUALLY the products are CO2 and H2O. At any rate, here is what you do. Yes, balance the equation. Then dHrxn = (n*dHf procuts) - (n*dHf reactants) Look up the dHf in your text or notes. Lots of places on the net also. By the way, ...
December 1, 2014

chemistry
Actually you didn't change the formula. The formula is M = mols/L and you just applied some algebra and rearranged it to mols = M x L. Note if you want L you would rearrange to L = mols/M
December 1, 2014

chemistry
You were going great then you got de-railed. Yes, mols = M x L BUT C (that's molarity) = mols/L and that's by definition (and the 6.0 is in the problem) so sticking that 0.04 in there is a no, no. mols = M x L = 6.0(from the problem) x 0.04 L = 0.24 mols Then grams ...
December 1, 2014

Chemistry
The 2.98 L is a false alarm and just extra data. Ignore it. So is the 4.90L. Ignore it too. You have how many mols Na3PO4. That's mols = grams/molar mass = ? So mols Na^+ = 3x mols Na3PO4. and mols PO4^3- = mols Na3PO4. I know students don't like to ignore data (it'...
December 1, 2014

chemistry
Just ignore the Ne. n = mols O2 = grams/molar mass = ? Then use PV = nRT, substitute and solve for pO2. Note: The partial pressure of Ne can be determined exactly the same way. The total pressure then is the sum of the two partial pressures OR you can add mols O2 to mols Ne, ...
December 1, 2014

Chemistry
That calculation looks ok to me.
December 1, 2014

Science
In science the idea is to think it through. So think about it. If you have a HYDRATED (do you know what that means?) ionic compound in a test tube and you HEATED it, wouldn't you expect that perhaps you might drive the water out of the hydrate (hydrated means it contains ...
December 1, 2014

Chemistry
In one word, no. You can calculate what you should get in the lab if you're doing a lab experiment. molar mass CuSO4.5H2O = 249.7 molar mass 5H2O = about 90 Then (90/249.7)*100 = ?
December 1, 2014

chemistry
mols NaNO3 needed = grams/molar mass. Using the coefficients in the balanced equation, convert mols NaNO3 to mols Pb(NO3)2. In this case that's mols NaNO3 x (1 mol Pb(NO3)2/2 mols NaNO3) = x mols NaNO3 x (1/2) = ?
December 1, 2014

Science IPS
I don't have the solubility tables nor a graph for that data; you must have that in your text/notes. If you will post that I can help you through it
December 1, 2014

chemistry
I suggest the following: 1. Write one post at a time. 2. Write posts in complete sentences. 3. Explain fully what you want and what you have or have done. 4. For pics etc why not look on Google yourself instead of us doing it. That way you get the pics you want or that make ...
December 1, 2014

chemistry
No. See your later post way up the page.
December 1, 2014

Chemistry
K2CO4 + BaCl2 ==> BaCrO4 + 2KCl mols K2CrO4 = M x L = approx = 0.02 mols BaCl2 = M x L = approx 0.01 The limiting reagent is BaCl2 so you will obtain 0.01 mols BaCrO4. g BaCrO4 = mols x molar mass = ?
December 1, 2014

chemistry
Use PV = nRT and solve for n = mols CO2. Then mols CO2 x (5 mols ZnO/2 mols CO2) = mols ZnO. Then convert mols ZnO to g. g = mols x molar mass. I would be somewhat concerned that CO2 is soluble in H2O, you haven't made a correction for the vapor pressure of H2O and you'...
December 1, 2014

Chemistry
There are 6.022E23 atoms H in 1 mol H atoms so the mass of 1 is (1/6.022E23 = ?).
December 1, 2014

Chemistry
This is a buffer problem with the slight complication of added HCl. The NH4I + NH3 is a buffer solution and the pH is determined by the Henderson-Hasselbalch equation. The complication is that HCl has been added to it so the thing to do is to calculate mols NH3 and subtract ...
December 1, 2014

Chemistry
You can modify the gas law of PV = nRT to P*molar mass = density*RT Substitute and solve for molar msss.
December 1, 2014

Chemistry
If K = 2.48E85, then note you didn't square or to the 4th power where you should and the complete equation would be 2.48E85 = (2x)^2(x)/(0.1-4x)^4 I see I omitted the square term for (NO)^2 in my first response. The units of x are mols/L = M so since you have 1.0L that ...
December 1, 2014

Chemistry
You didn't provide any of the data but my first impression is that you do not need to calculate anything in reverse. If you have delta G, then use dG = -RT*ln*K (solve for K) and set up the ICE box as ........S + 4NO ==> SO2 + 2N2O I...0.025..0.1.......0.....0 C.....-x...
December 1, 2014

Chemestry
density = mass/volume. Plug and chug. The units will be in g/mL.
December 1, 2014

chemistry
Crystals are solids. Usually crystalline means they are not amorphous; i.e., no crystal structure. If it is an ionic crystal it is a repeating row/column of positive ions and negative ions such as you find in NaCl, etc.
December 1, 2014

Chemistry
Now you're answer mooching. I showed you how to do this type question under Amanda. You get more help when you don't change screen names.
December 1, 2014

Chemistry
December 1, 2014

Chemistry
I would think that graph which you couldn't post would give you the wavelength. At any rate, the peak is at the lower wavelengths http://en.wikipedia.org/wiki/Black-body_radiation I found this at 5000 K which is close to 4500 K. This shows 5,000 K about 500 or 600 nm so my...
December 1, 2014

Chemistry
Temperature if you are talking about radiant heat and the composition and temperature if you are talking about spectral colors.
December 1, 2014

Chemistry
You can work this as a ratio/proportion because you know 2 mols NH3 (2*17=34 g NH3) will require +92.6 kJ (I just changed the sign because the question you asked is the reverse of the equation). (92.6 kJ/34g) = (? kJ/30.5 g) and solve for ? kJ. I do these this way. 92.6 kJ x (...
December 1, 2014

chemistry
meaning what? You want to know those numbers. Google them. I'm sure the web has them.
December 1, 2014

Chemistry 2
You didn't put a + charge on one of the Ag. It has to be on the left side 2H2O + MnO2 ==> MnO4^- + 3e + 4H^+ Ag^+ + e ==> Ag^+ Those are the two BALANCED half reactions. Multiply equation 1 by 1 and equation 2 by 3 and add them.
December 1, 2014

chemical equations
Aluminum + copper(II) chloride-> copper + aluminum chloride Al of course is aluminum. For coper(II) chloride it tells you the valence of Cu is +2. Chloride from the table I gave you is -1. All compounds are zero so you must make the + charges and the - charges add up to ...
December 1, 2014

chemical equations
The Roman numeral TELLS you the valence of the ion. Cu(I) means Cu^+1. Cu(II) means Cu^2+. Fe(III) tells you Fe^3+. Couldn't be easier than that. If it has no Roman numeral, look for it on the periodic table. If it is group I (or 1) (the left most column) it is +1. If in ...
December 1, 2014

chemical equations
I've helped you with one; now I see a bunch more of the same kind. What's the problem? What do you not understand about all of this. Do you the valences? Do you know how to guesstimate what the valences are? Do you know the polyatomic ions and their valaences?
December 1, 2014

chemical equations
Cu(NO3)2 + 2NaOH --> Cu(OH)2 + 2NaNO3
November 30, 2014

chem 200
specific heat ice = ? heat fusion water = ? specific heat H2O = ?
November 30, 2014

chemistry
November 30, 2014

Chemistry
This is a buffer problem and you must use the Henderson-Hasselbalch equation, The C9H7NHBr is the base. HBr is the acid.
November 30, 2014

chemistry
Why are you having trouble converting units? 1 m = 100 cm You have the conversion factor for m to nm. Have you ever seen one of these sodiujm vapor lamps? (I'm sure you have--They are yellow) f = frequency in Hz w = wavelength in m c = speed of light = 3E8 m/s Substitute ...
November 30, 2014

CHEM
Look up the reductions potentials and see which half cell reacts.
November 30, 2014

Chemistry
1. c won't do it. Count the charge. On the left there is 14+ (from 14H^+) + 2- (from Cr2O7^2-). On the right there is 2*3 = 6+. So you Add 6e to the left side. 2. b won't get it. In redox reactions electron gain is ALWAYS ALWAYS ALWAYS = electrons lost (if the equation...
November 30, 2014

Chemistry
Go to Google and type in"activity series". The metals at the top are the top reducing agent. Metal ions at the bottom are the top oxidizing agents.
November 30, 2014

chemistry-chemical equations
Cu + HNO3 ==> Cu(NO3)2 + NO2 + H2O Can you balance it? Cu goes from zero oxidation state on the left to +2 on the right. N (in HNO3) goes from +5 on the left to +4 on the right (in NO2)
November 30, 2014

Chemistry
I assume that is 1.5% NaCl w/v; if so then you have 1.5g NaCl/100 mL solution. That is the same as 15g NaCl/L solution and that has a molarity of 15/58.44 = approx 0.257. So you are diluting 0.257M NaCl from 1.0 mL to 100 mL. New molarity is M = 0.257 x (1 mL/100 mL) = ?M
November 30, 2014

chemistry
I would convert the measurements to cm; therefore, 300 cm x 275 cm x 250 cm = 2.06E7 cc. Convert cc to L = 2.05E7 cc x (1L/1000cc) = approx 2.06E4 L but you need to confirm all of these numbers and if needed with more accuracy. Then PV = nRT. Substitute 10.7 torr(converted to ...
November 30, 2014

Chemistry
I doubt I can help find such a video; however, in this case I don't know what I'm looking for. You didn't give the reaction or the chemical formulas.
November 30, 2014

Chemistry
%abundance Br79 is 50.54% Other isotope Br is 100-50.54 = 49.46% (78.904*0.5054) + (mass*0.4946) = 79.904 Solve for mass (other isotope) = ?
November 30, 2014

Chemistry (Titrations)
No, no, Bridget. That will not work in this case BECAUSE 1 mol Fe is not = 1 mol MnO4^-. 1. Write and balance the equation. Fe(NH4)2(SO4)2.6H2O + KMnO4 + H2SO4 ==> Fe2(SO4)3 + (NH4)2SO4 + K2SO4 + MnSO4 + H2O 2. Convert 0.2848g Fe(NH4)2(SO4)2.6H2O to mols. mol = grams/molar ...
November 30, 2014

Chemistry
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. mols Zn = grams/atomic mass = ? mols HCl = M x L = ? Using the coefficients in the balanced equation, convert mols Zn from above into mols H2. Do the same for mols HCl to ...
November 30, 2014

chemistry
mols Ca(OH)2 = M x L = ? Convert mols Ca(OH)2 to mols HClO4 using the coefficients in the balanced equation. Then M HClO4 = mols HClO4/L HClO4. You know mols and M HClO4, solve for L HClO4 and convert to mL.
November 30, 2014

Chemistry
a. Use PV = nRT and solve for n = number of mols O2. mols NO = grams/molar mass = ? Step 2. Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one ...
November 30, 2014

Biochem
I am unable from your post to know exactly what you're doing; however, in a hydrolysis of an ester you produce an alcohol and an acid. I would assume that the NaOH titration is to determine how much acid (total, not just HCl) has been produced. My best guess is that the ...
November 30, 2014

Chemistry
See your last post. You also need to convert 36F to C AND you must know the initial temperature of the water.
November 30, 2014

Chemistry
You need to look up the delta H solution of NH4NO3. It probably will be listedin kJ/mol. Convert 8g to mols and multply by delta H to find q for this problem. Then -q = (mass H2O X specific heat H2O x (Tfinal-Tinitial) Solve for Tf
November 30, 2014

introductory chemistry essentials
new M = 5.8M x (3.5/55) = ?
November 30, 2014

Chemistry
I can't read the values you have in the table because of the problem with spacing on this forum but here is what you do. dSo = (n*dS products) - (n*dS reactants) dH you have but it can be done the same way. After you have dSo rxn and dHo rxn, then' dGorxn - dH - TdS.
November 30, 2014

Chemistry
An ICE chart makes it easier, I think, to see. You have written the equation for the dissociation constant for the complex, the number given is a formation constant; therefore, I will rewrite it as (and note that the charge on the complex ion is 3- and not 3+.) The easier way ...
November 29, 2014

chemistry
% w/w (by mass) = (mass solute/mass solution)*100 = ? mass solute = 3.10g mass solution = 3.10g + 10.4g = ?
November 29, 2014

Chemistry
1.40 x (75.0/248) x [124/(124+165)] = ?M
November 29, 2014

Science
q1 for raising T of ice from -15 to zero C is q1 = mass ice x specific heat ice x (Tfinal-Tinitial)= ? (and I would work this as a separate step since you have all of the numbers and put that number for q1 below ) q2 to melt ice is q2 = mass ice x heat fusion = ? (Again, you ...
November 29, 2014

Science
What specific heat are you to use for liquid H2O?
November 29, 2014

chemistry
3.3% by mass means 3.3g Ag/100 g solution. The density of the solution is 1.02 g/mL so 3.8L (3,800 mL) has a mass of 1.02 g/mL x 3800 mL = 3876 g. (x g Ag/3876)*100 = 3.3 x g Ag = approx 128 g. That's an estimate.
November 29, 2014

Chemistry
The combustion of benzoic acid is C7H6O2 + 15/2 O2 ==> 7CO2 + 3H2O and since you want the enthalpy of FORMATION (and not combustion), reverse the equation (and the sign for dH combustion) to obtain 7CO2 + 3H2O ==> C7H6O2 dH = +3223.6 kJ Then dHrxn = (n*dHf benzoic acid...
November 29, 2014

Chemistry
Convert 1.60 mg K2Cr2O7 to mg C2H5OH as follows: mols K2Cr2O7 = 0.00160/molar mass K2Cr2O7. Using the coefficients in the balanced equation, convert mols K2Cr2O7 to mols C2H5OH. That's mols K2Cr2O7 from above x (3 mols C2H5OH/2 mols K2Cr2O7) = mols K2Cr2O7 from above x (3/...
November 29, 2014

chemistry
Use the Henderson-Hasselbalch equation. Sodium benzoate is the base and benzoic acid is the acid.
November 29, 2014