# Posts by DrBob222

Total # Posts: 54,515

**chemistry**

K^+ is 0.2M ......F^- + HOH ==> HF + OH^- I....0.2............0.....0 C.....-x............x.....x E...0.2-x...........x.....x Kb for F^- = (Kw/Ka for HF) = (x)(x)/(0.2-x) Solve for x = (HF) = (OH^-) You have OH^-, then Kw = (H^+)(OH^-) so you can solve for (H^+). All of ...

**Chemistry**

You don't. You are on the wrong path to the answer. .......2NO(g) + O2(g)---> 2NO2(g) I.......10.0....10.0........0 C.......-2p.....-p.........+2p E.....10-2p.....10-p........2p Substitute the E line into the Kp expression and solve for p and 2p.

**Chem**

I don't get it. The problem gives you specific instructions for working it. You do it step by step. If you need help please explain what you're having trouble with in detail. Show your work you've done and explain what you don't know to do next.

**Chemistry**

What's wrong with using (P1V1/T1) = (P2V2/T2) Remember to use T in kelvin. Convert C to K with K = 273 + C.

**Chemistry**

Here is an activity series table. Use it to answer the question. http://www.grandinetti.org/activity-series Here is how you do it. Any METAL will displace (react with) any METAL ION below it in the activity series. For example, Sn will react with Cu^2+ ion. Sn + CuCl2 ==> ...

**Chemistry**

Sorry but this forum does not allow us to draw.

**Chemistry**

I assume you mean ZnCl2. mols = M x L = ?

**chemistry**

1 ppm = 1 mg/L; therefore, 2500 ppm = 2500 mg/L. You want 250 mL; therefore, 2500 x 250/1000 = ? mg/250 mL. Now convert mg to grams and you have it.

**chemistry**

5% w/w H2SO4 means 5 g H2SO4/100 g solution. So you want 5g H2SO4 and 95 g H2O BUT be sure you add the H2SO4 to the H2O and not the other way around. That gives you 100 g solution. To make 400 mL just multiply everything by 4.

**Chemistry**

Did you work it or someone told you?

**Chemistry**

mols NH3 = grams NH3/molar mass NH3 = ? mols urea = mols NH3 x (1 mol urea/2 mols NH3) = mols NH3 x 1/2 = ? Then grams urea = mols urea x molar mass urea = ?

**Chemistry**

Technically the concentrations go in in molarity. I like to work in mols and since M = mols/L and the value for L in both numerator and denominator is the same the L cancel and one can use mols (actually I use millimols) directly. millimols HAc = 500 mL x 0.3M = 150 mmols...

**chemistry**

P1V1/T1 = P2V2/T2 Remember T must be in kelvin. K = 273 + C

**Chemistry**

See your other post.

**chemistry**

0.1 M x (3 mL/10 mL) = ?

**Chemistry**

Use PV = nRT Watch the units.

**CHEMISTRY**

SrCl2 => Sr^2+ + 2Cl^- m = mols solute/kg solvent If SrCl2 is 0.15 m then Sr^2+ must be 0.15 m and Cl^- must be 0.30 m

**Chemestry**

I don't see a question here but you need to learn how to spell chemistry.

**chemistry**

What, pray tell, is Cl2Cl2.

**Chemistry**

With good technique and good equipment, it is possible to get very close to 100%. Your theoretical yield is 0.52 and you obtained 0.52 for CO2 so that is 100% yield. If you had weighed to two more places it may (or may not) have been exactly 100%. If you use 44.01 for molar ...

**chemistry**

3.05 g - 1.94 = 1.11 g H2O abd 1.94 g CuSO4 mols H2O = grams H2O/molar mass H2O mols CuSO4 = grams CuSO4/molar mass CuSO4. Now fine the ratio of the two to each other. The easy way to do that is to divide the smaller number by itself (thereby making sure it is 1.00). Then ...

**Chemistry**

I think you're guessing (big time) but your guessing gene is pretty good. We don't do quizzes but I just confirmed your guess.

**Chem**

Heat lost by water + heat gained by Pt = 0 We call Pt 1; then H2O = 32 [mass H2O x 32 x (Tfinal-Tinitial)] + [mass Pt x 1 x (Tfinal-Tinitial) = 0 Substitute and solve for Tfinal.

**Chemistry**

mols CH3OH = grams/molar mass = ? 1 mol CH3OH contains 6.02E23 molecules. Then there are 4x that number of H atoms..

**Chemistry**

N = 14 O = 16 NO = 30. Therefore, 30 g NO will contain 6.02E23 molecules mols NO in 2.5E20 molecules = 2.5E20/6.02E23 = ? mols NO. 1 mol has a mass of 30 g so ? mols will have a mass of .....

**Chemistry**

Ge has an atomic mass of 72.6; therefore, 1 mol (72.6g) will contain 6.02E23 atoms of Ge. How many mols do you have? That's grams Ge/72.6 = ?

**Chemistry**

Do you have an equation for this. Are you sure you didn't go through an extra step with iodide ion?

**Chemistry**

See my question and response above.

**Chemistry**

You need to be careful how you write questions; I think you meant that 10 mL H2SO4 was diluted TO 250 mL with DW. I'll assume that's what you meant. 2NaOH + H2SO4 ==> Na2SO4 + 2H2O mols NaOH = M x L = 0.1 x 0.01 = 0.001. mols H2SO4 in that 25 mL aliquot = 1/2 x 0....

**Chemistry**

Sarah, I worked this for you below. It's a waste of our time to answer twice AND you posted under Lauren also. I assumed no water vapor in the flask.

**Chemistry**

This is a limiting reagent (LR) problem. mols Zn = 0.01 mols HCl = M x L = 0.03 x 1 = 0.03 mols H2 produced from Zn if we had all of the HCl we needed = 0.01 (from the 1 mol Zn gives 1 mol H2. molsl H2 produced from 0.03 mols HCl if we had all of the Zn we needed = 0.03 x 1/2...

**chemistry**

2NaOH + H2SO4 ==> Na2SO4 + 2H2O I think for (b) you means how many mols H2SO4 were used. All of ot was meeded. mols H2SO4 = M x L = ? (c) Using the coefficients in the balanced equation, convert mols H2SO4 you had to mols NaOH neutralized. (d) usually these problems want ...

**Chemistry**

See above under a difference screen name. Sarah, you continue to use difference screen names and I've asked you several times to use the same screen name. It really slows us down in getting the answers out to you AND it serves no useful purpose.

**Chemistry**

k = 0.693/2 min = ?. Then subsitute k into the below equation. ln (Ao/A) = kt Let Ao = 100, then A = 25. Substitute Ao, A, and k into the equation and solve for t in minutes.

**Chemistry (Urgent)**

First things first. Learn how to spell celsius. 1. The equilibrium equation is just the solubility in water. PbI2(s) ==> Pb^2+(aq) + 2I^-(aq) 2. Surely you know how to write the Ksp. 3. From the equation is (1), solve for the solubility. .....PbI2 ==> Pb^2+ + 2I^- I......

**Chemistry**

Use wavelength = h/mv. That's the De Broglie equation.

**Chemistry**

What's wrong with substituting into n = m/M? n = 140/28 = ?

**chemistry**

Al^3+(aq) + 3OH^-(aq) ==> Al(OH)3(s)

**homework**

NO. Why divide by 4. Didn't Scott say times 4 weeks. And what did you do with the 7 days per week.

**Chemistry**

[H2O2] before = 0.88M volume H2O2 before mix is 4 mL total volume after mix is 5 (4 from H2O2 + 1 from KI) so the H2O2 has been diluted from 4 to 5 mL.' 0.88 x 4/5 = ? KI before mix is 0.500. volume KI before mix is 1 mL. total volume after mix is 5 (1 from KI + 4 from ...

**chemistry**

Almost. You were asked to calculate the mass CO2 generated. The last line should read "0.5 g of carbon dioxide" 05 g/mol is not right. The unit /mol canceled in your previous step.

**Chemistry**

Note you have the wrong formula for fluorapatite. It is Ca5(PO4)3F. 2Ca5(PO4)3(OH) + SnF2 ==> 2Ca5(PO4)3F + SnO + H2O This is just like the problem I did for you last night. These stoichiometry problems are done the same way. A four step process. 1. Write and balance the ...

**chemistry**

Your full equation is ok. The half equations are Zn ==> Zn^2+ + 2electrons O2 + 4 electrons ==> 2O^2- For the full equation you multiply equation 1 by 2 and add to equation 2.

**Chemistry**

Ksp = (2x)^2(0.1) You can look up the Ksp for Ag2CrO4. It's approximately 9E-12 but you should use the value in your book's table of Ksp values. 9E-12 = 4x^2(0.1) 9E-12/0.1 = 4x^2 9E-11 = 4x^2 9E-12/4 = 2.25E-11 = x^2 x = sqrt 2.25E-11 = 4.47E-6 M In 2L that is 4.47E-6...

**chemistry**

I use this most often but it's a couple of equations put together. mL x M x (molar mass/1000) = grams You don't list mL in the problem and you must have that. M - 0.272. Molar mass acetylsalicyclic acid you can look up. Calculate grams. Then % ASA = (grams ASA/0.726 ...

**Chemistry**

dG = dGo + RTlnQ Write the equation. Determine the limiting reagent. Q = (NaNO3)^2/[Pb(NO2)] Determine the concentrations for Q and calculate. Post your work if you get stuck.

**Chemistry**

When dG is - (i.e., <0) the reaction is favored. Therefore, if dGo is >0 it must be + and that means the system is not favorable for a reaction. That is the equilibrium lies to the left. That takes care of 1 and 5. 2. Knowing that the equilibrium lies to the left means ...

**Chemistry**

This looks like a limiting reagent (LR) problem. SiO2 + 2C ==> SiC + CO2 mols SiO2 = grams/molar mass = approx 50/60.1 = approx 0.0832 mols C = 50/12 = approx 0.0416 mols SiC formed using just SiO2 = 1:1; therefore, 0.0832. mols SiC formed using just C = 0.0416 x 1/2 = 0....

**Homework**

You ask questions, we provide answers BUT the emphasis is on HELPING you and not doing all of your work. We try to teach you how to do it rather than doing it for you. It is not a place to type in your tests and make 100% score. Here are the tips you should do. Homework ...

**Chemistry**

Ag2CrO4 ==> 2Ag^+ + CrO4^2- Ksp = (Ag^)^2(CrO4^2-) mols CrO4^2- = mols x L = ? Substitute into Ksp expression and solve for (Ag^+). That is (AgNO3). Convert M to mols in the 2.0 L and convert to grams. g = mols x molar mass. Post your work if you get stuck.

**add on----Chemistry**

You have the two solubilities. You will need to calculate the ratio from that.

**Chemistry**

CaSO4 ===> Ca^2+ + SO4^2- For pure water, Ksp = (Ca^2+)[(SO4)^2-] So (Ca^2+) = x and [(SO4)^2-] = x. Solve for x and that is the solubility of CaSO4 in pure water. For the 0.1M Na2SO4 solution. Ksp is the same expression. (Ca^2+) = x [(SOr)^2-] = 0.1 + x. [Note: that is 0....

**chemisrty**

mols KOCl = 6.22g/90.55 = 0.0687 or 68.7 millimols. ......OCl^- + H^+ ==> HOCl I....68.7.....0........0 add..........16............. C...-16.....-16........16 E....52.7.....0.........16 You can substitute to find the pH of this solution is 8.05 Out of the buffer range for ...

**Chemistry**

mols N2 = 2.0/28 = about 0.07 (N2) = 0.07/1.3L = about 0.05 mols O2 = 3.0/32 = about 0.09 (O2) = a0.09/1.L = about 0.07 All of the above need to be recalculated more accurately. .......N2 + O2 ==> 2NO I....0.05..0.07......0 C.....-x...-x.......2x E..0.05-x..0.07-x....2x ...

**Chemistry**

Take a 100 g sample to give you 57.8g C, 3.6g H and 38.6 g O. Convert to mols. 57.8/12 = ? 3.6/1 = ? 38.6/16 = ? Now find the ratio of CHO to each other with the lowest being no smaller than 1.00. The easy way to do this is to divide the smaller number by itself, then divide ...

**Chemistry**

mols C = 7.961/44 = ? mols H = 1.164 x (2/18) = ? mols N = 1.551/30 = ? Now find the ratio in small whole numbers to one another. The easy way to do that is to divide the smallest number by itself (which gives you 1.000), then divide the other numbers by the same small number...

**Chem**

Cindy, Tiffany, et al. See your other posts. We prefer you stick to one screen name. Posting with more than one name often delays help.

**Chemistry**

Your Ksp for AgBr is eqn 1. Your Kf for Ag(S2O3^-2)^3- is eqn 2. Add equn 1 to eqn 2 to get AgBr(s) + 2(S2O3)^3- ==> Br^- + [Ag(S2O3)2]^3- Write the equilibrium expressiion for that which equals Ksp*Kf. Then substitute 0.02 for Br^- and 0.02 for [Ag(S2O3)]^3- and solve for...

**Chemistry**

You don't need the gas equation. Mg + 2HCl ==> H2 + MgCl2 mols Mg = grams/atomic mass = ? This could be a limiting reagent problem so watch that.

**CHEMISTRY**

mols Na2CO3 = grams/formal mass = ? L = 2.25 M Na2CO3 = mols/L. (Na^+) = ? ppm 1 ppm = 1 mg/L g Na^+ = 2.22E-4 x 2 = 4.44E-4 g/2.25 L = ?g/L. Convert to mg/L to obtain ppm. (CO3^2-) in ppm is 1/2 that.

**Chemistry**

1/A - 1/Ao = kt A = 0.0174 M Ao = 0.0693 M t = 489 s Solve for k.

**Chemistry**

mols SO2 = 38000/molar mass SO2 = ? mols H2SO4 produced = same as mols SO2 becaue 2 mol SO2 = 2 mol H2SO4. Then grams H2SO4 = mols H2SO4 x molar mass H2SO4 = ? g, then convert to kg. That's if the process were 100% efficiency. It isn't so that answer x 0.70 = the ...

**Chemistry**

How many mols do you need? That's M x L = mols. 0l00 x 0.558 L = ? Then grams = mols x molar mass = ? That's grams you need if the SrCl2 is 100%. It isn't so you need ?g from above/0.586 = ??

**Chemistry**

You got an error message because 1/0 is an indeterminate. I didn't know how to write 1/infinity but that's what it should have been and 1/infinity is 0. As to the other, I guess I don't know what the second excited state is. I assumed it was with the electron in n...

**Chemistry**

E = 2.180E-18(1/2^2 - 1/0)

**Chemistry**

The energy of the H atom in its ground state is -2.18E-18. Find the difference in the energy, change the negative sign to a positive sign then E = hf and solve for f = frequency. Is that something like 10^15 Hz? You should confirm that.

**Chemistry**

Would you look at your post and proof it please. That 10^20? Is that right?

**Chemistry**

a. T goes up. E goes up. b. T goes down ..... c. V goes down so P goes up which means more collisions. d. More mols Ne means higher P so .....

**Chemistry**

The problem says nothing about opening a valve so you have the same amount of gas you started with unless there's a leak somewhere.

**chemistry, science**

Please find the caps key and use it. I don't know what he is. Perhaps you meant He. mols = grams/molar mass

**chemistry**

How much H3PO4 is in the sample? That's 75.0 x 0.175 = approx 13g but you need to do it more accurately. How many mols is that? That's about 13/98 = approx 0.13. 2H3PO4 + 3Ba(OH)2 ==> Ba3(PO4)2 + 6H2O mols Ba(OH)2 = approx 0.13 mols H3PO4 x (3 mol Ba(OH)2/2 mol ...

**Chemistry**

mols KOH = grams/molar mass = ? Then M KOH = mols/L. You have mols from above and convert 200 mL to L.

**Chemistry**

E = 2.18E-18 J x (1/1 - 1/0) = 2.180E-18 J x (1 kJ/1000J) x 2.78 mol H atoms x 6.02E23 atoms/mol = ?

**Chemistry**

See your other question above.

**Chem**

delta T = i*Kb*m delta T is what you want. i for MgBr2 is the van't Hoff factor = 3. Kb you can look it up. It't approx 0.5 m = mols/kg solvent. You have g water. Convert to Kg. Then mols MgBr2 = grams/molar mas.

**Chemistry Help**

When equimolar concentrations of base and acid are used, the pH is ALWAYS the pKa. Why? pH = pKa + log (base)/(acid) But if base = acid, then base/acid = 1 and log 1 = 0 so pH = pKa. a is pKa = - log Ka b is pKb = -log Kb, then pKa + pKb = 14

**Chemistry**

You must assume this is a non-volatile compounds AND that it is not ionic. delta T = 100.78-100 = ? delta T = Kb*molality You know delta T, look up Kb (about 0.5), solve for m. m = mols/kg solvent. You kow m and kg waterk solve for mols. mol = grams/molar mass You know grams ...

**Chemistry**

3.65 g HCl in 1 dm3 H2O is 3.65/36.5 = 0.1 M HCl mols HCl = M x dm3 = 0.1 x 0.0227 dm3 = ? Na2CO3 x 2HCl ==> 2NaCl + H2O + H2O mols Na2CO3 = 1/2 mol HCl = ? and that's the mols Na2CO3 in 25 cc. Convert t mols Na2CO3 in 1 dm3 = ?mols Na2CO3 x 1000/25 = ? Convert that to ...

**Chemistry**

14 F. Convert to C. (F-32)*5/9 = C delta T = Kf*molality Substitute delta T and Kf. Solve for m. m = mols/kg solvent. You know m and kg H2O. Solve for mols. mols = grams/molar mass. You know mols and molar mass, solve for grams.

**Chemistry**

How many mols are in 5.00 g NaCl? That's mols = grams/molar mass = ? Then M = mols/L solution. 250 mL = 0.250 L.

**CHEMISTRY**

No. Oxidation is the loss of electrons. Reduction is the gain of electrons. Mg(s) with a zero oxidation number changes to Mg^2+ ion with a +2 oxidation number. From zero charge to +2 charge means it had to lose electrons so Mg is oxidized. Fe^2+ on the left changes to Fe(s) on...

**CHEMISTRY**

The equation ALWAYS tells you. 25 mols O2 = 18 mols H2O or 25:18

**Chemistry**

NH4OH initially = 0.1 mol NH4Cl initially = 0.1 mol When adding HCl the equation is ......NH3 + H^+ ==> NH4^+ When adding NaOH the equation is and I will do this one. ......NH4^+ OH^- ==> NH3 + H2O I.....0.1....0.......0.1....... add........0.02............. C....-.02..-...

**Chemistry**

How did you get the initial pH when you don't know how much base/acid you started with.

**chemistry**

NaOH + HCl ==> NaCl + H2O mols NaOH = M x L = ? mols HCl = mols NaOH since the equation tells you 1 mol NaOH = 1 mol HCl. Then M HCl = mols HCl/L HCl = ?

**Chemistry**

mols HCl = M x L = ? From the equation, mols Na2CO3 = 1/2 that. grams Na2CO3 = mols Na2CO3 x molar mass Na2CO3. %Na2CO3 = (grams Na2CO3/0.2638)*100 = ?

**chemistry**

How many mols do you want? That's mols = M x L = 0.1 x 0.1 = ? Then M HCl = mols/L. You know M = 4.0M and you know mols = ? from above, solve for L of the 4.0 M stuff. There is an easier way to do it. You should get the same answer eithr way. mL1 x M1 = mL2 x M2 100 x 0.1...

**Chemistry**

acetate is the base. You need NaOH with HAc (acetic acid). You don't need the HCl. pH = pKa + log (base)/(acid) 5.00 = pKa + log b/a. Look up pKa for HAc. Solve for b/a ratio. That's eqn 1. Equation 2 is b + a = 0.200 Solve those two equations simultaneously for b and ...

**chemistry**

Produce 25.4 g CuCl2 from what? Cu? amd Cl2? Cu + Cl2 == CuCl2 mols CuCl2 = 25.4 g/molar mols Cl2 = same as mols Cu. Then use PV = nRT, substitute mole Cl2 for n, substitute P and T (remember T must be in kelvin) and solve for V in Liters.

**Chemistry**

% w/w = % by mass = [(g solute)/(total grams)]*100 Total grams = grams solute + grams solvent

**Chemistry**

K2SO3 + 2HCl --> 2KCl + H2SO3 mols HCl = M x L = ? Using the coefficients in the balanced equation, convert mols HCl to mols K2SO3. Now convert mols K2SO3 to grams. g = mols x molar mass = ?

**chemistry**

M final solution = total mols/total liters. mols soln 1 = M x L = ? mols soln 2 = M x L = ? total mols = ? volume soln 1 = 0.276 L volume soln 2 = 0.125 L. total volume = ?

**Chemistry 101**

What about them?

**chemistry**

There is a chapter of work here. What do you not understand about each? How much do you know how to do?

**Chemistry**

What's your trouble with this. Concn products/concn reactants and the coefficients in the equation become exponents in the Kc expression.

**Chemistry**

So the polar solvent holds more tightly to the polar solute; the less polar solute has nothing to hold it back so it swishes through the TLC plate.

**Chemistry**

Yes, it does. Is dichloromethane a polar solvent?

**college chemistry**

Alex, I don't quite get it. The problem asks for how much carbon you use. Is that grams carbon. But the rest of the problem tell you to end up with grams CO2. Here is how you do it if you want grams CO2. 2660g/gal x 500 gal/year = approx 1E6 but you need to do it for a ...

**chemistry**

Take a 100 g sample which gives you K = 60 g C = 18.5 g N = 21.5 g Convert to mols. mol K = 60/39.1 = ? mol C = 18.5/12 = ? mol N = 21.5/14 = ? Now find the ratio of each element to each other. The easy way to do that is to divide the smallest number by itself (you know that ...

**chemistry**

See my response to your earlier CxHy question below. I've worked it in detail. I hope things work out for you very well.