Monday

July 28, 2014

July 28, 2014

Total # Posts: 43,208

**Chemistry**

I worked this for you a day or so ago. Find that and you have it. Or post your work here and I can show you where you're going wrong.

**Chemistry**

Use the Henderson-Hasselbalch equation. Post your work if you get stuck.

**chemistry**

Use PV = nRT and solve for n = number mols HCl. The M HCl = mols HCl/L solution. Then pH = -log(HCl)

**chemistry**

Xgas = n gas/total mols gas.

**chemistry**

Let's call these r1 and r2 for reagent 1 and reagent 2. So r1 is C7H6O3 and r2 is C4H6O3. You made a typo in the equation. You typed an = sign for a + sign. C7H603+C4H6O3=C9H8O4+HC2H3O2 a. Which reactant is limiting? mols r1 = grams/molar mass = estimated 1.45 mols r2 = g/...

**chemistry**

Read my response. Next to last sentence. "Now convert 3 mols CH3OH to grams. g = mols x molar mass. This is the theoretical yield (that is 100% yield)."

**chemistry**

Which is the limiting reagent? CO + 2H2 ==> CH3OH mols CO = gram/molar mass = approx 3 mols H2 = approx 53 Convert mols CO to mols CH3OH using the coefficients in the balanced equation. That's 3 x (1 mol CH3OH/1 mol CO) = approx 3 Do the same for H2 to CH3OH. That's...

**Chemistry URGENT**

The balancing is ok I think but the formula are not. NaCl +AgNO3 >>> NaNO3 + AgCl(s) This is ok. BaCl2 + H2SO4 >>> BaSO4 + Cl2H(s) hydrochloric acid is HCl, not Cl2H Na2CO3 + CaCl >>> 2NaCl(s) +CO3Ca calcium chloride is CaCl2 NaOH + HNO3 >>>...

**Chemistry URGENT**

You asked parts of this yesterday and Bob Pursley answered. pH = -log(H^+) or (H^+) = 10^-pH So pH = 12.2 means (H^+) = 10^-12.2 and pH = 4.6 means (H^+) = 10^-4.6 and I've written them this way because that's the question and we're told to express them this way. B...

**Chemistry**

millimols CO3^= = 100 x 0.6 = 60 millimols Ba^++ = 500 x 0.06 = 30 ......Ba^++(aq) + CO3^=(aq) --> BaCO3(s) I......30......................... added..............60............ C.....-30.........-30............+30 E.......0..........30.............30 So this results in a co...

**chemistry**

Use PV = nRT. If that 1700 c means degrees celsius, don't forget to convert that to kelvin. If you are using 8.31 for R, you must use P in kPa. 1 atmosphere = 101.325 kPa.

**Chemistry**

I can't see the photo but a d orbital has a value of 2 for ell (I can't write the script ell and the lower case l looks like (OK, ALMOST like) the number 1).(see 1 vs l) s orbitals have ell = 0 p orbitals have ell = 1 d orbitals have ell = 2

**Chemistry**

(1/wavelength) = R(1/9 - 1/25) The 9 is 3^2 and the 25 is 5^2. R = 1.0973E7

**Chemistry**

wavelength = h/mv m = mass in kg v = velocity in m/s

**Chemistry**

1.29 g/cc x 25 cc x 0.381

**chemistry**

I think I did this last night. I would look first at calculating how much (Ag^+) must be to ppt AgCl. You have Ksp for AgCl and you have Cl- from KCl, that allows you to determine exactly how much Ag^+ is needed to ppt Agcl. Anything over that and AgCl will ppt. The question a...

**Chemistry URGENT/DR BOB**

All of the above looks ok to me except for these two. NaHCO3(s) + HCl(ag) >>> NaCl +2CO +H20 NaHCO3 + HCl ==> NaCl + CO2 + H2O CaCO3(s) +HCl(ag) >>> CaCl +CO3H CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

**Chemistry**

It's one mole to one mol; therefore, how many mols do you have in 3.82 g SA? That's mols SA = 3.82g/molar mass SA. Then mols AA = mols SA and g AA = mols AA x molar mass AA

**chemistry**

P1V1 = P2V2

**SCIENCE PPPLLLLLSSSSS HELP!!!!!!!!!!**

Igneous rock becomes sediment first, then upon being deposited and compacted when other layers form on top it becomes sedimentary rock.

**Chemistry**

I'm not sure this has a solution since there is no way to know the mole ratio between H2 produced and mols M consumed. I have assumed +2 metal and 1mol M to 1 mol H2 ratio to make MCl2 but it may not be that at all. M + 2HCl ==> MCl2 + H2 PV = nRT P = Ptotal - pH2O = 75...

**chem**

Come on shadow. The q1 is to get you from ice at -33.5 to zero. The q2 is to get you from solid ice to liquid water. The q3 is to get you from zero C to 100 C. And q4 is to get you from liquid water at 100 to steam at 100. The problem asks for heat need to go from -33.5 to ste...

**chem**

heat needed to raise T of solid water from -33.5 to solid water at zero C. q1 = mass ice x specific heat solid ice x (Tfinal-Tinitia) where Tfinal is zero and Tinitial is -33.5 q2 = heat need to melt ice; i.e., change solid water at zero C to liquid water at zero C. q2 = mass ...

**chemistry**

If that's the only information available to you I would suggest that you subtract 0.265-0.247 to find how much the radius decreases from period 6 to period 5 and subtract that difference from Rb to make an educated guess at K that's down one more period.

**Chemistry**

Can't answer. The change in pH depends upon the kind of acid or base. Weak bases/weak acids don't change the same way as strong acids/strong bases.

**chemistry**

Your posting leaves a little to be desired. It's quite difficult to interpret. For this problem, use the Henderson-Hasselbalch. Substitute the numbers and solve for pH.

**science-chemistry**

KHP + NaOH ==> NaKP + H2O mols KHP = grams/molar mass mols NaOH = mols KHP (from the 1:1 ratio inthe balanced equation.) M NaOH = mols NaOH/L NaOH. Withthe vinegar you didn't provide enough data but it's done this way. mL vn x M vin = mL NaOH x M NaOH.

**Chemistry URGENT/DR BOB**

Write an equation to show how acetic acid reacts with water to produce ions in solution. C2H4O2 + H2O >>>>>>> H3O + C2H3O You need to show the charges on the ions; i.e., H3O^+ + C2H3O^- Write an equation for the neutralization of HCl and NaOH. HCl + NaOH &...

**chemistry**

You didn't list k1 and k2 and the numbers in my text probably don't agree with the numbers in you text.

**chemistry**

.....H2C2O4 ==> H^+ + HC2O4^- I....0.35M......0......0 C.....-x........x......x E....0.35-x.....x......x Substitute the E line into k1 and solve for x = (H^+) = (HC2O4^-) .......HC2O4^- ==> H^+ + C2O4^= Write the k2 expression. You've just calculated that H^+ = HC2O4...

**chemistry**

What's K?

**chemistry**

Use the second order rate equation of (1/A) - (1/Ao) = kt and solve for A at the allotted time. That will give you NO2 also. Convert to mols NO2 and mols N2O4 and use PV = nRT to solve for P. Post your work if you get stuck.

**chemistry**

1.1655 g/mL x 1000 mL x %NaOH x 1/40 = 4.37M. Solve for %.

**chemistry**

144.2 pm = 1.442E-8 cm and that's the RADIUS. volume of sphere is (4/3)*pi*r^3. See if that gets you off on the right foot?

**chemistry**

What is it you don't understand about this problem. We can take it from there. Post any work you've done on this problem.

**chemistry**

H2O(l) + H2O(l) + heat ==> H3O^+(aq) + OH^-(aq) So if you add heat that forces the equilibrium to the right (to use up the added heat) which means a larger percent ionization as well as an increase in Kw at higher temperatures.

**chemistry**

This looks pretty obvious to me but tell me what you don't understand about it. How many particles do you expect from the three materials.

**chemistry**

I would suggest you go about it this way. First, you know Cl from KCl concn. Using Ksp for AgCl, and the KCl concentration, calculate (Ag^+) needed to ppt AgCl. Then using Kf for Ag(S2O3)^3-, calculate the Ag^+ in that solution. Compare that with Ag^+ from the Ksp. Post your w...

**cheMYSTERY**

I worked this problem earlier for nitrate. Mg is done the same way. It's a little more complicated for Ag because AgCl is a ppt. The total volume is 95 + 20 = 115 mL. After mixing, (AgNO3) = 0.8 x (95/115) = about 0.66 (MgCl2) = 0.2 x (20/115) = about 0.035 but that is not...

**CHEMISTRY**

rate S2O3^2- is 1/2*0.002M/s rate I2 consumption is 0.002 M/s

**chemistry**

WO3 + 3H2 ==> W + 3H2O mols WO3 = grams/molar mass = approx 0.17 Using the coefficients in the balanced equation, convert mols WO3 to mols H2O. That's ?mols WO3 x (3 mols H2O/1 mol WO3) = ?mols WO3 x 3 = xx mols H2O. Then convert xx mols H2O to grams. g = mols x molar m...

**CHEMISTRY!**

millimoles MgCl2 = 20 x 0.200 = 4.0 millimoles AgNO3 = 95 x 0.8 = 7.6 .........2AgNO3 + MgCl2 ==> 2AgCl + Mg(NO3)2 The nitrate ions never enter any reaction so they are at the end what they were initially. They were 0.8M but that's been diluted by 95/(95+20) That is 0.8...

**Chemistry**

a. 3-methylcyclohexene alkene. double bond b. Cis-3-octene alkene. double bond c. 4-methyl-2-heptanone ketone. R2-C=O d. 3-bromobenzaldehyde aldehyde. RCHO

**chemistry**

a. PV = nRT and n = grams/molar mass b. (V1/T1) = (V2/T2) 3. (P1V1/T1) = (P2V2/T2) 4. This is a redox equation. This link will tell all about how to do these. http://www.chemteam.info/Redox/Redox.html

**chemistry**

base = b acid = a First you must determine the (acid) and (base). You do that by solving two equations simultaneously. b+a = 0.1M is equation 1 pH = pKa + log (acid)/(base) and substitute like this from the problem 5.00 = 4.76 + log (b/a) is equation 2. Solve the two for a and...

**Chemistry pH**

You need to do an ICE table. The reaction is CH3NH2 + HOH ==> CH3NH3^+ + OH^-

**chemistry**

2C4H10 + 13O2 ==> 8CO2 + 10H2O mols O2 = grams/molar mass Using the coefficients in the balanced equation, convert mols O2 to mols CO2. Now convert mols CO2 to grams. g = mols x molar mass

**Chemistry**

I have corrected the equation you wrote. .......CaF2(s) = Ca2+(aq)+2F-(aq) I......solid.....0.........0 C......solid.....x.........2x E......solid.....x.........2x Substitute the E line into the Ksp expression and solve for x = solubility in mols/L. grams = mols in 1 L x molar...

**Chemistry**

mols Ca(OH)2 = 1.78/74.09 = about 0.024 M = mols/L = about 0.024M ........Ca(OH)2 ==> Ca^2+ + 2OH^- I.......solid........0........0 C.......solid........x........2x E.......solid........x........2x Ksp = (Ca^2+)(OH^-)^2 = ? (x)(2x)^2 = (0.024)(2*0.024)^2 = Ksp = ? You need ...

**Chemistry**

This problem is worded so funny that I'm afraid to give an answer. Electrons generated at the anode(an electrode) go through an outside circuit to the cathode(another electrode), then through the electrolyte and back to the anode. From that description perhaps you can figu...

**Chemistry pH**

HC2H3O2 = HAc = acetic acid C2H3O2^- = Ac^- = acetate ion ...........HAc --> H^+ + Ac^- I........0.30......0....0.95 C..........-x......x......x E........0.30-x....x....0.95+x Ka = (H^+)(Ac^-)/(HAc) 1.8E-5 = (x)(0.95+x)/(0.30-x) Solve for x and convert to pH.

**Chemistry**

Mg + Cu(NO3)2 ==> Mg(NO3)2 + Cu mols Mg = 145.9/24.3 = 6.003 The equation is 1 mol to 1 mol; therefore, you will have 6.003 mol Mg(NO3)2 formed. Molar mass I have is 148.31 to give g Mg(NO3)2 =890.29g at 100% yield. 890.29 x 0.3860 = 343.65g actual yield. You're allowed...

**analytical chem**

You have 1 millimole HCl added to 1 millimole NaOH which exactly neutralizes each other to give you NaCl and H2O for which the pH is 7.0 (ignoring the activity coefficient corrections).

**Inorganic Chemistry**

2MnO4^- + 4H2O + 3C2O4^2- ==>6CO2 + 2MnO2 + 8OH^-

**Chemistry**

0.318mol/4.15L = 0.0766M .......PCl5 ==> PCl3 + Cl2 I...0.0766.......0.......0 C........-x......x.......x E...0.0766-x.....x.......x Substitute the E line into the Kc expression and solve for x and evaluate 0.0766-x/

**chemistry**

So Qsp = (Ag^+)^2*(CO3^2-) (Ag^+) = 0.013M x (1000/1005) = ? (CO3^2-) = 0.17M x (5/1005) = ?

**Chemistry**

As the size of the molecule increases and the molar mass increases the van der Waals forces increase the boiling point of these molecules.

**Chemistry**

I don't know how you obtained C2H5. Here is what you do. You must first convert everything to C, H, and find the difference in order to find Pb. g C = grams CO2 x (12/44) = ? g H = grams H2O x (2/18) = ? g Pb = 3.279 - gC - gH Then mols C = gC/12 = ? mols H = gH/1 = ? mols...

**To Drbob222**

Much closer but not quite. I think part of the problem is the 1.386. I think that should be 4.158. k = 0.693/20 = 0.03465 ln(125/N) = 0.3465*120 ln(125/N) = 4.158 Take the antilog of both sides. antilog ln(125/N) = 125/N antilog 4.158 = 63.94 So 125/N = 63.94 and N = 125/63.94...

**To Drbob222**

No. If the sample starts out at 125 ug and it loses half of the sample every 20 seconds, how can it have more than it started with at the end of 2 minutes. It MUST be less than 125 ug.

**To Drbob222**

No, you aren't following the math. The half life is 20 seconds. k = 0.693/t1/2 = 0.693/20 = 0.03465. Then ln(120 ug/N) = 0.03465*120 Solve for N. By the way, that ln (which you don't have anywhere in your work) in front of the (No/N) is not for looks. That means to tak...

**To Drbob222**

You follow the math. k = 0.693/t1/2 Then ln(No/N) = kt No = 120 ug N = solve for this k = from above t = 120 sec. First, you substitute the half life of this material which is listed as 20 seconds and that allows you to solve for k. You need k to solve the equation below it in...

**Chemistry**

Use PV = nRT and solve for n = number of mols. Then n = grams/molar mass. You know n and molar mass, solve for grasm. Another way to do it is to recognize that at STP 1 mol of a gas occupies 22.4 L. So mol = 31/22.4 and that x molar mass = grams.

**Chemistry**

2H2O2 ==> 2H2O + O2 mols O2 gas = 0.460L/22.4 = ? Using the coefficients in the balanced equation, convert mols O2 gas to mols H2O2. Now convert mols H2O2 to grams. g = mols x molar mass.

**Pre Chem**

(P1V1/T1) = (P2V2/T2)

**chemistry - (Dr. Bob222)**

On re-reading the problem I note that the problem lists NaOH as 0.1M (and I jumped to the conclusion that was 0.1N). It certainly IS 0.1N but the point is that I assumed the prof was working in normality and not molarity. Therefore, this may not have been a normality problem i...

**chemistry - (Dr. Bob222)**

I can do this step by step if you wish but I remember one neat formula that wraps all of it up together. mL x N x mew = grams. 24mL x 0.1M x mew = 0.125 Solve for mew. 0.05208 so eq. wt. will be 52.08 (but that's too many significant figures). I looked up the molar mass an...

**Radioactivity or science**

k = 0.693/t1/2 Then ln(No/N) = kt No = 120 ug N = solve for this k = from above t = 120 sec.

**Radioactivity**

I obtained about 11,000 (11,466 but that's too many significant figures) years, too, so you probably did it right.

**chemistry**

q = mass I2 x heat fusion but I should note that I2 sublimes rather than melts.

**chemistry**

What's the concn of the iv solution?

**Chemistry**

q = mass x heat fusion. 8.19 kJ = 24.6 x heat fusion. Remember this is + in going from solid to liquid.

**chemistry**

The first one is right but you don't need parentheses around (SO4). ZnSO4 is sufficient. The second one is right if you intended @ to be 2 for HCl. And the h of HCl should be capitalized.

**chemistry urgent**

First, determine q for the reaction. That is 25700 J x (45.5/molar mass NH4NO3) = approx 15,000 J. Then q = [mass H2O x specific heat H2O x (Tfinal-Tinitial) q from above mass H2O = 125g specific heat H2O given Solve for Tf Ti is given

**chemistry**

I think D and B have definite answers. D. 3 mols left; 3 right. dS zero so TdS zero. that + a - dH gives dG - so we go with products favored. B. 4 mols left; 3 right. dS is - which makes -TdS + and that added to + dH means +dG so reactants favored. A and B are up for grabs. He...

**Chemistry**

I agree with your approach with the coefficients of 2 and reverse the C==>D. But aren't you supposed to add? You're adding the equations to get the final equation; therefore, you add dH and dS. And where did the 722 come from? I believe you are confusing this with d...

**Chemistry**

mols NaOH = grams/molar mass M NaOH = mols/L solution (NaOH) = M NaOH pOH = -log(OH-) pOH + pH = pKw = 14. You know pKw and pOH, solve for pH.

**Chemistry**

No, it's 1.48 M times 0.042 L and divide by 049, solve and convert to mL. However, you can save some time if you do it this way. 1.42 x 42 mL and divide by 0.49. That gives the answer in mL and there is no conversion to do.

**Chemistry**

HCl + NaOH ==> NaCl + H2O mols NaOH = M x L = ? mols HCl = mols NaOH (from the coefficients in the balanced equation.) Then M HCl = mols HCl/L HCl. You know mols and M HCl, solve for L and convert to mL.

**Chemistry**

See your post below.

**Chemistry**

3NaOH + H3PO4 ==> 3H2O + Na3PO4 mols H3PO4 = M x L = ? mols NaOH = 3*mols NaOH from the coefficients in the balanced equation. Then M NaOH = mols NaOH/L NaOH.

**Chemistry**

dHorxn = (n*dHo formation) - (n*dHo proucts).

**Chemistry**

I'm not exactly sure I understand the question. dH is enthalpy. dS is entropy. dG = dH - TdS So dG is negative for spontaneity. I think this is just an algebra exercise but check me out on that. If dH is + that's the opposite for getting a negative number for dG so tha...

**Chemistry**

dH for THIS reaction (just the 1.045g) is q = 50g x 4.184 x (Tfinal-Tinitial). MOST of the time these problems want the answer expressed in kJ/mol. If that is the case here that would dH in kJ/mol = q from above x (molar mass CaO/1.05g) x (1 kJ/1000 J) = ? aprox 82 kJ/mol

**chemistry**

2HNO2 + Sr(OH)2 ==> Sr(NO2)2 + 2H2O

**chemistry**

You do this just like the other one. Post it if you have a specific question about what you don't understand.

**chemistry**

C3H8 + 5O2 ==> 3CO2 + 4H2O Use PV = nRT and solve for n = number of mols C3H8 Using the coefficients in the balanced equation, convert mols C3H8 to mols of CO2.

**chemistry Equations**

Not all right but the right idea. When a strong base is added (such as NaOH) it reacts with the weak acid (CH3COOH) of the buffer. CH3COOH + NaOH ==> CH3COONa + H2O When an strong acid (such as HCl) is added, it reacts with the weak base of the buffer (CH3COO^-). CH3COO^- +...

**chemistry**

Right. You must use 483 mL of the 10M HCl stock solution. So how much water must be added? That's 2100-483 = 1617 mL. I suspect that since the problem uses exponent notation as well as L, that the answer must be expressed the same way. I'm assuming you are allowed 2 si...

**chemistry**

No. I converted the 2.100E0 L to mL so the answer on the other side will come out with mL. But the 10 mol/L stock solution has the same units as the 2.3 mol/L which makes things ok. You could change them, of course, but if you change one you must change the other. But it won&#...

**chemistry**

mL x M = mL x M 2100 x 2.3 = ?mL x 10

**Bio**

http://www.dadamo.com/typebase4/depictor5.pl?233

**oops--Chemistry**

I omitted a very important word in one of the last sentences. What I wrote was, "For gaseous reactions just remember that increased causes a shift to the side with fewer moles (so as to occupy the smaller volume)." I should have written, "For gaseous reactions j...

**Chemistry**

Le Chatelier's Principle is all about which way the reaction will shift when we mess around with a system in equilibrium. In plain English it states that a system in equilibrium will try to undo what we do to it. For a system such as A + B ==> C + D + heat. If we add A ...

**Chemistry**

I obtained 2.4 also.

**Chemistry - check answer**

I agree with both.

**Bio/Chem**

both. The 3% is 3g/100 mL (w/v) and the 1.5% is only 1.5g/100 mL so you have changed the concentration. Obviously there is less H2O2/100 mL in the 1.5% solution. But you have changed the percent too. Obviously, from 3% to 1.5%.

**Chemistry**

1:1 ratio for what? Mg + 2HCl ==> H2 + Mg. Use PV = nRT and solve for n = number of mols at the conditions listed. For P you want dry H2 and not wet H2 (it was collected over H2O and will be wet). Ptotal = 745 pH2O = look up vapor pressure H2O at 19C. Pdry H2 gas = 745-vapo...

**Chemistry**

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

**Chemistry**

Amen! and some don't make any sense at all.

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