Wednesday

July 23, 2014

July 23, 2014

Total # Posts: 43,165

**Chemistry pH**

pH = -log(H^+) pH = 3.72; (H^+) = approx 2E-4 but that's an approximation. You should do this more accurately as well as all of the calculations that follow. %ion = [(H^+)/Msample]*100 ...........HA ==> H^+ + A^- I.........0.6.....0......0 C...........-x....x......x E.....

**Chemistry**

This forum is not designed for graphing, drawings, spacings, etc etc. What the problem wants you to do is to graph P vs 1/V So the first point will be 1 atm for P and 1/150 mL for volume The second point will be 1.15 atm for P and 1/130 mL for V etc

**Chem**

In ==> In^3+ + 3e tough eh! Cd^2+ + 2e ==> Cd

**Chemistry**

Your set up is right (except for calling torr psi). So your error must be in the math part.

**Chemistry**

No, 519 cc isn't correct either. Post your work and I'll find the error.

**Chemistry**

I didn't work 1 but 2 can't be right. 775 is Ptotal so how can it be pO2?

**Chemistry**

1. (P1V1/T1) = (P2V2/T2) Remember T must be in kelvin. 2. Ptotal = pO2 + pH2O Substitute and solve for pO2.

**Honors chemistry**

q = mass x heat fusion

**CHEM- need help ASAP**

I worked this for Jon earlier. Post any follow up questions here. http://www.jiskha.com/display.cgi?id=1397681371

**Chemistry**

mols Na = grams/molar mass Using the coefficients in the balanced equation, convert mols Na to mols Cl2. Now convert mols Cl2 to liters remembering that mols x 22.4L/mol = ?L

**Chemistry**

Let's write ethylamine as BNH2. If the pH is 11.87 the pOH is 14-11.87 = pOH of 2.13 and 2.13 = -log(OH^-) so (OH^- is approx 7E-3M but you need to that more accurately as well as all of the other calculations that follow. Then .......BNH2 + HOH ==> BNH3^+ + OH^- I........

**Chemistry**

See your Kb ethylamine post. Same procedure. Post your work if you get stuck.

**chemistry**

The solution is acidic at the equivalence point due to the hydrolysis of the cation of the weak base.

**Chemistry**

The CN^- is hydrolyzed. ...........CN^- + HOH ==> HCN + OH^- I.........0.8..............0.....0 C..........-x..............x.....x E.........0.8-x............x.....x Kb for CN^- = (Kw/Ka for HCN) = (x)(x)/(0.8-x) and solve for x = (OH^-) = (HCN) This give you the HCN. Conve...

**chemistry**

I'm not exactly sure what you want but I would do this for the balanced chemial equation. For the thermo equation you want this without the 2 coefficient and you want to add the dH = -129 kJ/mol 2NaHCO3 ==> Na2CO3 + CO2 + H2O NaHCO3 ==> 1/2 Na2CO3 + 1/2 CO2 + 1/2 H2O...

**chemisrty**

Looks like not enough information is there but it can be done. The easy way to understand it is to assume some convenient concn for the weak acid and the NaOH and work it out. The easy way to solve it is to do it by reasoning. You can go through the numbers if you wish but the...

**chemistry**

Use the HH equation. The way I read the problem you have a solution that is 0.2M in propionic acid and another solution 0.2M in sodium propionate. pKa = -log Ka = about 4.89 but you should confirm that and adjust the figure to your liking. 4.65 = 4.89 + log (base/acid) b/a = 0...

**chemistry**

pN2 = XN2*Ptotal

**College Chemistry (DrBob222)**

k = 0.693/t1/2 ln(No/N) = kt No = any convenient number but I would use 100 to represent 100%. Then if it decreases by 73% that leaves 27% so N = 27 k from above t = ? Solve for t in seconds if you use half life in seconds.

**chemistry - (Dr. Bob222)**

I obtained 1.51 which is essentially the same as your answer. Did they give a reaction? The formula actually is 1/A - 1/Ao = akt so the a could change things but usually is not a number like 2/3. Must be a wrong answer. If you find to the contrary please be sure and post so I ...

**chemistry - (Dr. Bob222)**

[1/(A)] - [1/(Ao)] = kt k is given in seconds; therefore, change 2.5 minutes to seconds.

**CHEMISTRY HELP !!!!**

What's with this "show your work"? What work? No information is furnished. You can look up the Ka value in this table. http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm

**Chemistry - Science (Dr. Bob222)**

Can't you simply plug in k1 and k2 along with T1 and T2 and solve for activation energy with the Arrhenius equation?

**Chemistry - Science (Dr. Bob222)**

See your other post above.

**Chemistry**

2KNO3 ==> 2KNO2 + O2 The equation above is not necessary for solution of the problem. Use PV = nRT and solve for n.

**Chemistry**

I obtained, using the equilibrium as 2HI ==> H2 + I2 as K = (H2)(I2)/(HI)^2 = (0.1)(0.1)/(0.7)^2 = 0.0204 which would be 1/0.0204 = 49 for the way you suggested.

**Chemistry**

I have made mistakes like that but in this case I didn't. It could be either way. Which ever way you do it will work out. If you assume you started with HI, then 2HI ==> H2 + I2 and the EQUILIBRIUM MIXTURE will be as given. Then you work out Kc for that reaction. You ad...

**Chemistry**

First, convert mols to M. (HI) = 7/10 = 0.7M (I2) = 1/10 = 0.1M (H2) = 1/10 = 0.1M Next calculate the Kc. 2HI ==>H2 + I2 Kc = (H2)(I2)/(HI)^2 Kc = (0.1)(0.1)/(0.7)^2 = approx 0.02 Then do and ICE chart. ............2HI ==> H2 + I2 I...........0.7....0.1...0.1 add...........

**Chemistry**

Actually all of these can be worked using the table of reduction potentials but I will give that process last and show shortcuts for metals vs metal ions and for halogens vs halogen ions. The metals are done by looking at the activity series of metals. Here is a simplified cha...

**Chemistry**

I assume you performed some experiment and you are to use your observations to help answers this question. What were your observations?

**Chemistry**

0.9% w/w NaCl in water means 0.9g NaCl/100 g solution. So you put 0.9 g NaCl in 99.1g H2O and shake till dissolved. You want to make 200 g of this so double those numbers.

**Chemistry**

Why can't you simply substitute into the formula given in the problem?

**Chemistry**

This post has gotten pretty far down the list and I may not see it by tomorrow. If you're still confused please post a new question at the top of the page.

**Chemistry**

I think your part way there but I don't think you grasp the problem. And n =2 is the last shell of F. First, the electronic configuration of F is 1s2 2s2 2p5. You can see from this, and the problem restates that, that there are 9 electrons in the F atom. Thr problem is ask...

**Chemistry**

n is the principal quantum number and in general terms is the old Bohr shell. l (ell) is the azimuthal quantum number and describes the ellipticity (how elliptical) of the orbit. ml describes how the orbit is arranged in space ms describes the spin of the electron; i.e., clock...

**Chemistry**

You need to learn to do these. Here are the rules. n can be any whole positive number beginning with 1, 2, 3, etc. l(ell) can be any positive whole number beginning with zero but can't be larger than n-1 ml can have values from -ell to + ell. ms may have either of two valu...

**Chemistry**

I think so although no number is placed on it in that table you looked at. Most might say iodine since there is so little At around.

**Chemistry**

http://en.wikipedia.org/wiki/Template:Periodic_table_%28electron_affinities%29

**Chemistry**

For a 1 tesla field the Larmor frequency is freq = 28.025 GHz. http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nmr.html

**CHEm**

Use PV = nRT and solve for n = number of mols. Then n = grams/molar mass. You know n and molar mass, solve for grams. I would convert cubic meters to L.

**Chem**

PV = nRT and solve for n = number of mols. Then n = grams/molar mass and solve for molar mass.

**Chem**

mass = volume x density

**Radioactivity**

k = 0.693/t1/2 Substitute k into the equation below. ln(No/N) = kt No = 120 N = ? k from above t = 120 seconds (since the half life is given in sec you will have a seconds terms for k so you must convert 2 minutes to seconds). Solve for N. See the link below for the definition...

**Chemistry**

What do you mean you don't know where to start. This is just a case of substitute and solve. 7.7E24 = Kc(0.08205*298)^(4-2) Solve for Kc.

**Chemistry Help**

Coulombs = amperes x seconds = approx 3000. 107 g Ag will be deposited for every 96,485 coulombs. So Ag deposited is 107g Ag x (approx 3000/96,485) = ? g Ag.

**Chemistry**

A. The reaction for A is just the reverse of the original; therefore, the new Kp(that is K'p) for the new rxn will be 1/Kp B. B is just 1/2 the original; therefore, the new Kp (that's K"p) will be sqrt(Kp) For any reaction of A ==> B, then Kp = pB/pA. So when K...

**CHEM**

What is Ag? What's the other number for?

**CHEM**

Your post isn't very clear. Do you mean this is a concentration cell with one half cell of 1E-2M and the other 3.2E-2. If so then Ecell = -(0.05916/1)log (dilute/concd)

**chemistry**

mols sample = 0.05/114 = approx (but you do it for the real number) = 0.00044 q = mass H2O x specific heat H2O x delta T. q = 200g x 1 cal/g*C x 5.7 q = approx 1100 calories for 0.05g or 1100/0.05 = approx 23000 calories 23,000 calories x (1 kcal/1000 cal) = 23 kcal = 23 nutri...

**chemistry**

3,5,5 look just like 1 and 2. 6 and 7 are done by working backward. delta T = i*Kf or Kb *m You're given delta T, solve for m. Then m = mols/kg solvent. Solve for mols Then mols = grams/molar mass. You're given mols and grams; solve for molar mass. i = 1 for both 6 and 7

**chemistry**

2. Find the freezing point of a solution composed of 50g IBr in 120 g of water. Assume 100% ionization. KBr = k1+Br1-. I assume you made a typo and IBr really is KBr. mols KBr = grams/molar mass. Solve for mols. molality = m = mols/kg solvent Solve for m delta T = i*Kf*m i fo...

**chemistry**

1. Find the boiling point of a solution composed of 110g of HgCl2 (a non-ionizing solute) in 175g of water. mols HgCl2 = grams/molar mass Substitute and solve for mols. molality = m = mols/Kg solvent Substitute and solve for m delta T = i*Kb*m Your post says i = 1 Kb = 1.51 S...

**chemistry**

dGo = -nFEo Solve for n. That will give you the change in electrons.

**Chemistry**

I differ slightly (84.82) probably because you rounded to 1.62 while I kept everything in the calculator until the final answer. But to three significant figures I would round the answer to 83.8 (probably krypton).

**Chemistry**

(rate1/rate2) = sqrt(M2/M1) Substitute and solve for molar mass.

**science**

Oh my! Twisted minds. I'm still smarting over carbon dioxide and water having the same physical properties.

**science**

Generally days are shorter in the winter than in the winter.

**Science**

A is the correct answer. Water is H2O and carbon dioxide is CO2. They have different physical properties (water is a liquid while CO2 is a gas at room temperature.) Both have a set ratio of atoms as shown by the formulas I've written. Water and CO2 have different chemical ...

**Science**

And a c'mon to Kalie, too. What are physical properties? What are chemical properties.

**Science**

I don't agree with that answer. Carbon dioxide is a gas; water is a liquid. And you think they have the same physical properties? C'mon.

**chem**

You're right about Pgas = X*Ptotal but I don't think you understand the equation I wrote. Dalton's Law of partial pressures tells us that the total pressure inside a container is the sum of the individual gas pressures. So I wrote the equation Ptotal = pO2 + pH2O. ...

**chem**

The temperature is used to look in a table of T vs vapor pressure. It tells you the vapor pressure of the water (in this case 25.81 mm Hg) at 26.4 C. I don't agree with your answer. And why in the world would you change mm and torr to atm if the problem didn't ask for ...

**chem**

Ptotal = pO2 + pH2O

**chemistry**

You can calculate delta S for the system and delta H for the system at 25 degrees C by using dHo formation and dSo formation and dHo rxn = (n*dH products) - (n*dH reactants) dSo rxn = (n*dH products) - (n*dH reactants). What's the temperature. dS surrounds = -dH/T dS syste...

**CHEMISTRY HELP !!!!**

dGsystem < 0 for any spontaneous process.

**Chemistry**

pH = -log(H^+) Substitute and solve for pH.

**chemistry**

M of the diluted stuff is 1.14 g/mL x 1000 mL x 0.20 x 1/98 = 2.33M M of the strong stuff is 1.84 x 1000 x 0.98 x 1/98 = 18.4M Then use the dilution formula. mLA x MA = mLB x MB 100 x 2.33 = mLB x 18.4M Solve for mL B. I obtained approx 13 mL

**chemistry**

mols glyc = grams/molar mass mols H2O = grams/molar mass Xgly = mols gly/total mols XH2O = mols H2O/total mols.

**chemistry**

PV = nRT

**Chemistry**

I'm not sure what you want for #1. We can't draw structures on this forum. For #2, the Cl outside the coordination sphere is the ionic Cl, therefore Cl- + Ag^+ ==> AgNO3. The two coordinated Cl atoms within the sphere are not ionic and do not react with AgNO3.

**chem**

1. mols NH3 = M x L = ? Using the coefficients in the balanced eqauation, convert mols NH3 to mols H2SO4. Now M H2SO4 = mols/L, You know mols and M, solve for L and convert to mL if needed. 2. You have mols NH3. Using the coefficients convert mols NH3 to mols (NH4)2SO4. Now co...

**Chemistry**

Write and balance the equation. Convert 400.0g C3H8 to mols. mol = grams/molar mass Using the coefficients in the balanced equation, convert mols C3H8 to mols or the product. Now convert mols of the product to volume. mols x 22.4 L/mol = Liters.

**chemistry**

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

**chem**

Density = 1.025 g/mL; therefore, 100 mL has a mass of 1.025g/mL x 100 mL = 102.5g. 100 mL H2O should weigh 100 g; therefore, there are 2.5 g vinegar in the sample and that is 2.5%. This assumes that the volumes of vinegar and H2O are additive.

**Chem**

q = mass Al x heat fusion Note that mass is in grams and heat fusion is in kJ/mol; therefore, you must convert g Al to mol or heat fusion from kJ/mol to kJ/g

**chemistry**

What is the "mass you calculated" refer to?

**chemistry**

What reaction? What change in % did you calculate? I don't get the question?

**chemistry**

mols F^- = 0.0025g/atomic mass F and since that is is 1 L, that is the concn in mols/L. Evaluate Qsp and compare with Ksp.

**chemistry**

(P1V1/T1) = (P2V2/T2) P1 = 1 V1 = 0.550 T1 = 273 P2 = ? V2 = 7.50E-2L T2 = 273 + 20

**Chemistry**

No and partly because I may have led you astray. The numbers I gave you are for 1L of the solution at that concn. You only want 500 mL of that (which I missed on first reading) so you need only 1/2 of your final figure or 5.346/2 = ?.

**Chemistry**

Li3PO4. You want (Li^+) = 0.1385M You want to make this out of Li3PO4. But since Li3PO4 furnishes 3 Li ions for every molecule of Li3PO4, you only need 1/3 that number or 0.1385/3 = ? mols Li3PO4. Then g Li3PO4 = mols x molar mass.

**Chemistry**

CH3COOH + NaOH ==> CH3COONa + H2O mols NaOH = M x L = ? mols CH3COOH = mols NaOH (because of the 1:1 coefficients in the balanced equation.) g CH3COOH = mols CH3COOH x molar mass CH3COOH. Then % CH3COOH = (g CH3COOH/mass sample)*100 = ?

**Chemistry**

This is a limiting reagent (LR) problem. You know that when an amount for BOTH reactants is given. Zn + 2HCl ==> H2 + ZnCl2 mols Zn = grams/molar mass. mols HCl = M x L = ? Using the coefficients in the balanced equation, convert mols Zn to mols H2. Do the same for mols HCl...

**Chemistry**

The Eo voltage in tables may changes as a result of acid or basic conditions; however, there are no H or OH ions in the half reactions you've written so you need not worry about that.

**Chemistry**

If that is the smallest Ecell, then turn those half cells into a reaction. Arrange the half cells so they produce a positive voltage. 2Eu^2+ + Sn^2+ ==> 2Eu^3+ + Sn

**Chemistry**

2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq). Ecell = Eocell - (0.5916/2*log Q and I know that confuses you but don't be. Plug in the numbers Eo and n and for Q plug in concns/pressures(in atm) as follows: Q = [(H^)^2*(Fe^2+)^2]/[(Fe^3+)^2*pH2] I write it this way because I'...

**Chemistry**

See your other post. delta e wrong at 6.

**General Chemistry**

3e not 6.

**Chemistry**

mols MgCl2 = grams/molar mass = ? M MgCl2 = mols/L solution = ? Since there are two mols Cl in 1 mol MgCl2, then M Cl = twice that of MgCl2.

**Chemistry**

An equation like this is confusing because I^-(on the left) is going to I2 (on the right) but it then dissolves in KI to make KI3 so let's simplify it by taking the KI out of it, balance it, then put the KI back in. 2I^- --> I2 I is -2 on left and 0 on right. Balance wi...

**chemistry help~~~~ Please!!!**

k = 0.693/t1/2 Then ln(No/N) = kt. No = 25 N = 1 k from above. t = solve for this.

**Chemistry-need help ASAP**

I'll get you started but I don't believe you can't do any of it. I don't know what the problem is asking for when it says "difference in pH". Difference in pH from WHAT? 15 mL of 3M NaOH added to 500 mL. (NaOH) = 3M x (15/515) = ? (Note: This assumes ...

**Chemistry-need help ASAP**

How much of this do you know how to do?

**Chemistry**

HAc + NaOH ==> NaAc + H2O First determine the M of the HAc. mL x M = mL x M 22.5 x .21 = 50x (I worked this part for you at your original post yesterday. Apparently you never went back to look at it.) M HAc = approx 0.9 but you need to do it more accurately. Now divide the ...

**Chemistry**

I responded earlier with this. How much do you know how to do?

**Chemistry**

Here us a site that tells you the details. http://www.chemteam.info/Redox/Redox.html I don't do them that way. Here is how I do the CrO4^-. You can add the states. CrO4^2- ==> Cr^3+ Cr is +6 on left and +3 on right. Add 3e to left. CrO4^2- +3e ==> Cr^3+ Count the cha...

**chemistry**

Did you square 1.75 or multiply it by 2. I think you did the latter; you should have squared it.

**Chemistry**

Mn is +7 in MnO4^-. It is +2 in Mn^2+. To balance the change in electrons we have MnO4^- + 5e ==> Mn^2+. I do it that way and make this the first step of the process of balancing. Some say there is a shorter way and you do this the LAST step and not the first. To do it that...

**Chemistry**

MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O

**chemistry 2**

HAc + NaOH ==> NaAc + H2O First, determine the molarity of the HAc. mL x M = mL x M 22.5 x 0.21 = 50*x x = 0.0945M Now divide the titration curve into four sections. a. at the beginning. b. between initial and eq point c. eq pt d. after eq pt Which of these do you have trou...

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