Saturday
June 25, 2016

Posts by DrBob222

Total # Posts: 52,572

Chemistry
What does this mean? "Name each compound by ionic, hydrogen, or molecular compound."
March 6, 2016

Chemistry
Dalton's Law of Partial Pressure. Ptotal = 1.023 atm from problem. pH2O = 42.2 mm Hg from problem. Depending upon how the answer is to be displayed (mm Hg or atm),convert one of the above into the other unit and use Dalton's law. Ptotla = pH2 + pH2O You know Ptotal and...
March 6, 2016

Chemistry
HC2H3O2 = HAc. [C2H3O2]^- = Ac^- It is easier to work in millimoles which I will do. Technically it isn't proper to use mols or millimols in the HH equation; however, since M = millimoles/mL and you have mmols/mL in numerator and denominator of the HH equation, the answer ...
March 6, 2016

Chemistry
See your other post above.
March 6, 2016

chemistry
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O mols H2SO4 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH. Then g NaOH = mols NaOH x molar mass NaOH = ?
March 6, 2016

chemistry12_1
N2 is 28 x 0.80 = approx 23 O2 is 32 x 0.20 = approx 6 Total approx 29
March 6, 2016

chemistry
Ca(HCO3)2 + 2HCl ==> CaCl2 + 2CO2 + 2H2O
March 6, 2016

chemistry
That are these? What are the units? H20=-228.6 NO= 86.69 N2H4= 159.3 02 = 0 If this is delta G, then dGrxn = (n*dGproducts)-(n*dG reactants) Then dGrxn = -RTlnK YOu know dGrxn, R, T, solve for K.
March 5, 2016

chemistry
You need to learn how to type the arrow. --> or ==> See your post under Lia. Same process. Post your work if you get stuck.
March 5, 2016

@Lia--chemistry
My guess is that's the red cabbage indicator question.
March 5, 2016

Chemisrty
HClO4 is strong acid. pH VERY low. HBrO is a weak acid. pH higher than HClO4 but still lower than 7. NH3 is a weak base. pH slightly more than 7 (maybe 8.5-10). Ca(OH)2 is strong base. Greater than NH3 but less than KOH. KOH is a very strong base and is VERY much higher than ...
March 5, 2016

chemistry12_2016
See your other post.
March 5, 2016

chemistry12
CaCO3 ==> CaO + CO2 mols CaCO3 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO. That's 1:1 so mols CaCO3 = mols CaO. Then g CaO = mols CaO x molar mass CaO.
March 5, 2016

chemistry
See your other posts with different screen names.
March 5, 2016

chemistry
You're asking the same question and apparently not doing much work of your own. Yes, it is a simple lab. No, it isn't a simpler answer. I searched Google, which you could do too, and finally came up with this equation. Red cabbage has a flavenoid in it. The particular ...
March 5, 2016

Chemistry
I may have missed something here but I don't think equilibrium ever stops working. If the equilibrium is disturbed, it MUST redirect itself so as to reach equilibrium again. And that will continue as long as something is changing.
March 5, 2016

Chemistry
What were you titrating? HOAc? What were you titrating with? NaOH or similar? Take the volume of base you added, add 20 mL to that and that is the final volume. The OAc^- then will be (in molarity) mols HOAc you started with divided by the final volume in L.
March 4, 2016

Chemestry
mols acetic acid (HAc) = grams/molar mass = ? mols KOH = mols HAc (look at the coefficients in the balanced equation. M KOH = mols KOH/L KOH. You know mols and M, solve for L and convert to mL if needed.
March 4, 2016

Chem
Are you assuming all form compounds with Pb^4+ Pb(CN)4 Pb(SO4)2 Pb(ClO3)4 Pb(H2PO4)4
March 4, 2016

chemistry
You're making it harder than it is. Mo(NO3)2.5H2O. 1 mole salt to 5 mols H2O. CaSO4.2H2O 1 mol salt to 2 mols H2O
March 4, 2016

Chemistry
Frankly, I don't know what you did in the red cabbage experiment. Without knowing what you did it's tough to answer any questions about it. That's why I gave you a link. If you google it you can find a lot of information.
March 4, 2016

Chemistry
You've done the hard part. (Na2SO4.6H2O) = 1.2 M. Now there is 1 sulfate/1 Na2SO4 so (SO4^2-) must the same as (Na2SO4). Right? Right. Now there are 2 Na/1 Na2SO4 so (Na^+) = 2 x (Na2SO4) = 2*1.2 = 2.4. Right?
March 4, 2016

Chemistry 101
See your other post.
March 4, 2016

Chemistry
mols = M x L = ? Then mols = g/molar mass. You know mols and molar mass, solve for grams.
March 4, 2016

Chemistry
It seems obvious to me that if you put more molecules together there must be a better and increasing chance that they will bump into each other. Right?
March 4, 2016

chemistry
http://www.webexhibits.org/causesofcolor/7G.html
March 4, 2016

Chemistry
rate1 = k1*(Q1)^a*(X1)^b where a and be are the exponents for the orders and (Q) and (X1) are the concentrations for rate 1. rate2 = k2*(Q2)^a*(X2)^b Then rate1......(k1*(Q1)^a*(X)^b ------------------------ rate2......(k2)*(Q2)^a*(X)^b Pick trial 1 and 2 since (X) is the same...
March 4, 2016

Chemistry
We're at a loss to draw diagrams on this forum. I'm sure you can find appropriate diagrams along with the appropriate explanations on Google.
March 4, 2016

Chemistry
The PE starts at a higher point than at the end, the difference is the amount of energy emitted.
March 4, 2016

Chemistry
It makes no difference: you didn't tell us what you did or even what the experiment was.
March 4, 2016

Chemistry
Convert g CO2 to g C. 0.320 g CO2 x (12/44) = ? Convert g H2O to g H. 0.0874 x (2/18) = ? Then g O = 0.214 - g C - g H. % C = [g C/0.214]*100 = ? % O = [g O/0.214]*100 = ? % H = [g H/0.214]*100 = ? Now take 100 g sample which will give you those percentages as g C, H, O. ...
March 3, 2016

chemistry
2) (V1/T1) = (V2/T2). Remember T must be in kelvin. The problem gives no initial volume except V. You can do either of two things. a. Assume any number for V1; then V2 will be twice that number. b. Just call the initial volume V1, then V2 will be 2V1. 5). Same procedure as 2) ...
March 3, 2016

Chemistry
I think you're looking for this. Just add the two equations you multiplied to get the electrons equal. 2In + 3Cd^2+ ==> 2In^3+ + 3Cd
March 3, 2016

AP CHEM
I answered a question quite similar to this a couple of days ago. There is no BaCl2 in solution. There are Ba ions and chloride ions and there are more chloride ions than barium ions. But to try to calculate a M of an ionic compound in which the positive barium ions and the ...
March 3, 2016

chemistry
q1 = heat needed to change phase of ice from solid at zero C to liquid at zero C is q1 = mass ice x heat fusion = ? q2 = heat needed to raise the T from zero C to 85 C. q2 = mass H2O x specific heat H2O x (Tfinal-Tinitial) = ? q total = q1 + q2
March 3, 2016

Chemistry
new mols = 0.178 + 0.094 = ? Then PV = nRT. Remember T must be in kelvin.
March 2, 2016

Chemistry
I would put it slightly differently. Molality is mols per kg SOLVENT ir mols/kg solvent. M usually stands for molarity and not mols. m usually stands for molality. I usually write m = mols/kg solvent.
March 2, 2016

Chemistry
Keep this procedure. It will come in handy working about 90 of the stoichiometry problems. 1. Write and balance the equation. 2Na + O2 ==> Na2O 2. Convert grams (usually given) to mols which is mols = grams/molar mass. In this case mols are given in the problem so this step...
March 2, 2016

Chemistry
Why didn't you tell us what you used as the limiting reagent? I could have found the error. AgNO3 + NaCl ==> AgCl + NaNO3 mols AgNO3 = M x L = 0.0576 mols NaCl = M x L = 0.064 So AgNO3 will form 0.0576 mols AgCl. NaCl will form 0.064 molg AgCl. The correct answer in ...
March 2, 2016

Chemistry
Let's call lactic acid HL. .........HL ==> H^+ + L^- I.....0.027M....0.....0 C.......-x......x.....x E....0.027-x....x.....x Look up Ka for Lactic acid and substitute the E line above into the Ka expression and solve for x = (H^+), then convert that to pH. Remember that...
March 2, 2016

Chem
I think you are looking for N-H and N-N bonds as sigma bonds.
March 2, 2016

Chemistry
The problem asks to show the states, also. The equation is correct. 2H2S(g) + 3O2(g) --> 2SO2(g) + 2H2O(g) Note the problem TELLS you the water is vapor so you know to write it as a gas and not a liquid.
March 2, 2016

Chemistry
https://answers.yahoo.com/question/index?qid=20110323110631AA6R4NF
March 2, 2016

Chemistry
Mg is oxidized to Mg^2+. Mg ==> Mg^2+ + 2e Cu^2+ is reduced to Cu(s) metal. Cu^2+ + 2e ==> Cu
March 1, 2016

Chemistry
I don't get the "or reacted". I assume you want the volume of H2 corrected to STP. Look up the vapor pressure of water at 10 C. Ptotal = pH2O + pH2 105.5 = vap pressure H2O + pH2 and solve for pH2. Then PV = nRT and solve for V. Remember to use the same units for...
March 1, 2016

Chemistry
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) You are looking for q. mass is given, you need to look up specific heat H2O (4.184 J/g*C) and you have Tfinal and Tinitial. Solve for q.
March 1, 2016

Chem
I thought I answered this yesterday.
March 1, 2016

chemistry
acetic acid is HAc. .........HAc ==> H^+ + Ac^- I..l....0.31.....0......0 C........-x......x......x E......0.31-x....x......x Substitute the E line into the Ka expression and solve for x. Then % ion = [(x)/0.31]*100 = ?
March 1, 2016

chemistry
See your post above.
March 1, 2016

Chemistry
You have it balanced except for the 8H2O. Just add 8H2O on the end. Na2SO4+Ba(OH)2*8H2O=BaSO4+2NaOH + 8H2O g Na2SO4 = 0.853 x 0.95 = ? g Ba(OH)2*8H2O = 1.707 x 0.87 = ? Convert to mols. mols Na2SO4 = grams/molar mass = ? mols Ba(OH)2*8H2O = grams/mola mass = ? Using the ...
March 1, 2016

Chemistry
This is not a very good problem because not all of the Ba(OH)2 is used AND because there is no BaCl2 formed. All of the Ba^2+ and all of the Cl^- is there that you started with; i.e., no actual BaCl2 is formed. So I don't understand what the problem wants since there is no...
March 1, 2016

Chemistry
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O mols NaOH = M x L = ? mols H2SO4 = 1/2 x mols NaOH (Look at the coefficients in the balanced equation.) Then M H2SO4 = mols H2SO4/L H2SO4 = ?
March 1, 2016

Chemistry
Cr(NO3)3(aq) + 3NaOH(aq) ==> Cr(OH)3(s) + 3NaNO3(aq) Now you convert that balanced molecular equation into a balanced ionic equation. Cr^3+(aq) + 3(NO3^-)(aq) + 3Na^+(aq) + 3OH^-(aq) ==> Cr(OH)3(s) + 3Na^+(aq) + 3NO3^-(aq) and here are the rules you follow. 1. Gases ...
March 1, 2016

Chemistry
I don't think it is. Single replacement reactions are the type AX + Y = YX + A The reaction above is AX + Y = AY + YX
March 1, 2016

Chemistry
.............HA ==> H^+ + A^- I...........0.1.....0.....0 C...........-x......x.....x E.........0.1-x.....x.....x The problem tells you pH = 2.44. Convert that to (H^+) by pH = -log(H^+) and substitute the x value in for (H^+) and (A^-). Evaluate 0.1-x and substitute that ...
March 1, 2016

Chemistry (stoichiometry)
2C6H6 + 15O2 ==> 12CO2 + 6H2O 22 mols C6H6 x (12 mols CO2/2 mols C6H6) = 22 x 12/2 = ? mols H2O = grams/molar mass = ? Then ?mols H2O x (12 mols O2/6 mols H2O) = ? mols H2O x 12/6 = xx
March 1, 2016

Chemistry- Dr.Bob
The 80 x specific heat ice x 15 = heat absorbed in changing the T of ice at -15 to zero C. Then you melt the ice and convert solid ice to liquid water with 80 x heat fusion ice. Then you combine the moving T to zero with melting and see if the 250 g water at 85 is enough to ...
March 1, 2016

Chemistry-Dr.Bob 222
That doesn't help me. I don't remember if I answered this or not and if I did what I said. If I answered, copy that and explain what you don't understand.
March 1, 2016

@pradeeo---Chemistry
I don't believe H2C2O4 has an equivalent weight of 63. The molar mass is 90 and it is diprotic; therefore, the equivalent weight is 45.
March 1, 2016

Chemistry
A common ion is just what you think it is. For example, a solution of NaNO3 and KNo3 has a common ion; i.e., the NO3^-. AgCl saturated solution with AgBr (or NaCl)O has a common ion. A buffer basically is a weak acid and its salt (such as acetic acid and sodium acetate) or a ...
February 29, 2016

Chemistry
The rxn is Fe + Cu^2+ ==> Cu + Fe^2+ so the electrons flow from Fe to Cu^2+. The anode is where oxidation occurs. Since the Fe is going into solution and the Cu^2+ is coming out of solution, the Cu side has too many nitrate ions.
February 29, 2016

Chemistry
Zn, Pb, Mg, Al, Ca will reduce Ag^+ to Ag BUT they also will reduce Sn^2+. See this site. http://www.csudh.edu/oliver/chemdata/data-e.htm
February 29, 2016

Chemistry (Food)
The answer is A IF that is 4 kcal for carb and 9 kcal for gat. Those periods after the ) sign bother me. That must be 4 and not .4.
February 29, 2016

Chemistry
I would do this. Note N changes, Br changes BUT some of the Br^- are 1- on the left AND 1- on the right so they don't change. When this happens balance the redox part then add enough Br^- in the final equation for those Br^- that don't change. 6e + 2N^3+ ==> N2 3H2O...
February 29, 2016

Chemistry... again
A polar bond has a charge separation while a non-polar bonds does not. That is, a Cl-Cl bond has equal pull on the electrons for each atom, each attracts those outside electrons equally, neither atom wins, and the electrons are distributed evenly. For the HCl bond, however, ...
February 29, 2016

Chemistry
The easiest to pull electrons away from is the largest.
February 29, 2016

Chemistry
acetic acid = HAc ......HAc + H2O ==> H3O^+ + Ac^- I....0.1......0......0 C....-x.......x......x E...0.1-x.....x......x Substitute the E line into the Ka expression and solve for H3O^+
February 29, 2016

Chemistry
See our other post.
February 29, 2016

chemistry
1. How many moles do you need? That's mols = M x L = ? Then convert to grams. mols = g/molar mass. You know mols and molar mass, solve for grams. 2. Use the dilution formula. mL1 x M1 = mL2 x M2 mL1 x 0.5 = 100 x 0.1 Solve f r mL1.
February 29, 2016

Some Chemistry Questions
1 and 2 are wrong. 3 is ok.
February 29, 2016

Chemistry
acetic acid = HAc NaC2H3O2 is NaAc mols NaAc = grams/molar mass = ? Then mols NaAc/0.250 = ?M mols HAc = molsL. mols = M HAc x L HAc volume = 0.250 L M = mols/L.
February 29, 2016

Chemistry
What is that set up. If can see what you're talking about I can walk you through each step.
February 29, 2016

Chemistry
........HA ==> H^+ + Ac^- I....0.095.....0.....0 C......-x......x.....x E...0.095-x....x.....x The problem tells you that the pH is 5.42 which means you can calculate (H^+) from pH = -log(H^+) and that is x in the above. Then (Ac^-) is the same and you can calculate 0.095-x...
February 29, 2016

Chemistry
That looks ok to me.
February 29, 2016

Chemistry
AgCl(s) ==> Ag^+ + Cl^-. Write Ksp for this. Ag^+ + 2NH3 ==> [Ag(NH3)2]^+ Write Kformation for this one. Overall is is AgCl + 2NH3 ==> [Ag(NH3)2]^+ + Cl^- Wrote Keq expression for this. Then if you will multiply Ksp*Kformation you will get the overall equilibrium ...
February 29, 2016

Chemistry
H2SO4 + 2NaOH --> 2H2O + Na2SO4 mols H2SO4 = M x L = ? Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH. Then M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L and convert to mL.
February 29, 2016

Chemistry
The limiting reagent controls the amount of product(s) formed.
February 29, 2016

cHEMISTRY
2AgC2H3O2 + MgCl2 ==> 2AgCl + Mg(C2H3O2)2 mols MgCl2 = grams/molar mass = ? Using the coefficients in the balanced equation convert mols MgCl2 to mols AgCl. Then convert mols AgCl to mass. g AgCl = mols AgCl x molar mass AgCl.
February 29, 2016

Chemistry
.........2CH4 ==> C2H2 + 3H2 I ......0.087......0......0 C........-2x.......x......3x E.....0.087-2x.....x......3x The problem tells you that 3x = 0.012 which allows you to calculate x and 0.087-2x (as well as 3x). Substitute those values into the Keq expression and ...
February 28, 2016

chemistry
First, I wouldn't start that way. Convert torr to atm; i.e., 310/760 = pi = ? pi = MRT. You know pi, R (0.0820), and T (in kelvin please). Solve for M = molarity. Then M = mols/L. You know M and L (or can convert to L), solve for mols. Then mols = grams/molar mass. You ...
February 28, 2016

Chemistry
So delta T = 0-(-3.55) = ? delta T = Kf*m You know delta T and Kf, solve for m = molality
February 28, 2016

Chemistry
mols KCl = grams KCl/molar mass KCl = ? Then M = mols/L solution.
February 27, 2016

Chemistry
I don't think anyone here is likely to do all of your work for you; however, if you will tell us exactly, and in detail, what you don't understand, perhaps we can help.
February 27, 2016

analytical chemistry
mL1 x M1 = mL2 x M2 20 x 0.1 = mL2 x 0.02 Solve for mL2 which gives you the FINAL volume. Subtract the final - initial to find amount to be added. Technically, that will not be right since volumes are not additive but for dilute solutions like this the difference is minuscule.
February 27, 2016

Chemistry
Zn + 2HCl ==> ZnCl2 + H2 mols Zn = grams/atomikc mass = ? Using the coefficients in the balanced equation, convert mols Zn to mols H2. Since this is a 1:1 reaction, mols H2 = mols Zn. Then use PV = nRT to solve for V at the conditions listed. Remember T must be in kelvin ...
February 27, 2016

Chemistry
Do you have a reaction equation? One source gives Fe2O3 as the product; another gives Fe3O4 as the product. The answer to this question depends upon which product is formed. I will assume Fe2O3 and go from there. 2Fe + 3H2O ==> Fe2O3 + 3H2 Use PV = nRT and solve for n = ...
February 27, 2016

chemistry
mols Fe2O3 = grams Fe2O3/molar mass Fe2O3 Then there are 2 mols Fe for every 1 mol Fe2O3.
February 26, 2016

Chemistry
n*lambda = 2d sin angle.
February 26, 2016

Chemistry
+7 K = +1 O = -2 and 4*-2 = -8 What must Cl be for KClO4 to be zero. Remember that all compounds or elements in the free state are zero total. +1 + ? + (-8) = 0
February 26, 2016

Chemistry
I disagree with Nerd. This is a limiting reagent (LR) problem. N2 + 3H2 ==> 2NH3 a. Convert 2 L N2 to mols NH3 produced if we had all of the H2 needed. That is 2 L N2 x (2 mols NH3/1 mol N2) = 2 x 2 = 4 L NH3 b. Convert 2 L H2 to mols NH3 produced if we had all of the N2 ...
February 26, 2016

chemistry
Note the correct spelling of Celsius. .........N2O4 <-> 2NO2 I........1.680.....1.220 C.........-p.......+2p E........1.680-p...2p The problem tells you that 2p = 0.550. You can calculate p from that. That allows you to evaluate 1.680-p Then substitute these values into ...
February 26, 2016

CHE 2B
1. Use the HH equation. pH = pKa + log (KNO2)/(HNO2) 2. Without values of pKa any thing I calculate won't agree with your calculations. Here is what you do. .......HNO2 + OH^- ==> NO2^- + H2O I......12.5....0........9.25 add............2............ C......-2.....-2...
February 26, 2016

chemistry
2C2H2 + 5O2 ==> 4CO2 + 2H2O Use PV = nRT for O2. Plug in 155 atm for P, V = 5.5L, Assume any value for T(in kelvin) and use your chosen value for T throughout. Solve for n = number of mols O2. Knowing you need that many mols O2, use the coefficients in the balanced equation...
February 26, 2016

Chemistry
Why don't you use sodium hydrogen carbonate for the name. Bob Pursley gave you the right procedure and the right numbers. That 84 should be for NaHCO3.
February 25, 2016

Chemistry
79 x 0.225 =
February 25, 2016

Chemistry
C7H16 + 11O2 ==> 7CO2 + 8H2O mols CO2 = grams/molar mass = 37.5/44 = approx 0.85 Convert mols CO2 to mols C7H16 using the coefficients in the balanced equation. approx 0.85 x (1 mol C6H16/8 mols CO2) = approx 0.85 x 1/8 = 0.1 mol C7H16 g C7H16 = mols C7H16 x molar mass ...
February 25, 2016

Chemistry
See your previous post.
February 25, 2016

Chemistry
5. Consider the following reaction: CO2 (g) + H2 (g) CO (g) + H2O (g) Calculate the value of the equilibrium constant, Kc, for the above system, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapour were present in a 2.00 L reaction ...
February 25, 2016

Chemistry
You should post each question separately. Some won't answer posts that too long; i.e., consists of several questions. Calculate the pH of a NaOH solution of concentration 0.5M (Kw = 1.0 x 10-14 mol2dm-6 at 25oC) (H^+)(OH^-) = Kw. Substitute and solve for (H^+) [note that...
February 25, 2016

Chemistry
You want to compare Ksp with Qsp. (AgNO3) = 1E-4 x (200/1100) = ?M (KCl) = 1E-6 x (900/1100) = ?M Qsp = (Ag^+)(Cl^-) = ? If Qsp > Ksp a ppt forms. If Qsp < Ks no ppt forms.
February 25, 2016

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