Monday

April 21, 2014

April 21, 2014

Total # Posts: 41,847

**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.

**Chemistry**

Scroll down on this site to see various classifications. http://en.wikipedia.org/wiki/Hard_water

**Chemistry**

27.9g = mass watch glass + dry xtals -24.5g = mass MT watch glass ---------- ??? = mass dry xtals.

**Chemistry**

How large is the cup? What's the concentration of the acid?

**oops Damon---Chemistry**

Na is 23 not 11; therefore, molar mass NaCl = about 58.5 mols NaCl = grams/molar mass = 1.2/58.5 = about 0.02 mols. M = mols/L = about 0.02 mols/0.2 L = ?

**Chemistry**

Na2CO3 + CaCl2 ==> CaCO3 + 2NaCl millimols Na2CO3 = mL x M = estimated 8 mmols CaCl2 = estimated 12.5 So you will form 8 mmols (0.008 mols) CaCO3.

**Chemistry**

1.2% is the absolute error. 17.1% is the relative error.

**Chemistry**

Mostly carbonates, bicarbonates, strong bases, phosphates (and the other phosphates;i.e., HPO4 and H2PO4) etc. You can read more about it here. http://en.wikipedia.org/wiki/Alkalinity

**Chemistry**

There is no question here.

**Chemistry**

25.687g = mass beaker + sample -25.683g = mass MT beaker ---------- 0.004g = mass TDS TDS = 0.004/50 mL. You convert it to whatever units you want.

**Chemistry**

I see no question here.

**Chemistry**

CaCO3. You should know the table of solubilities. Here is a simplified version. http://www.csudh.edu/oliver/chemdata/solrules.htm

**chemistry**

dH for H2O moving from 7.3C to 25C is dH = mass H2O x specific heat H2O x (Tfinal-Tinitial) dH for vaporization at 25C is dH = mass H2O x Hvap. Add the two together for the total.

**Chemistry**

K' = 1/K

**Chemistry**

A. With increased pressure volume decreases so the molecules have less space to move around in. S decreases. T is same so that doesn't enter into the problem. B. T and P remains same. The only difference is liquid vs gas. As a gas there is MUCH more volume to move around a...

**Chemistry need help**

Sorry, I forgot about that. dGo = -nFEo

**Chemistry need help**

Both can't be the cathode. OK. Zn --> Zn^2+ is the half reaction in the problem. What you looked up is the standard reduction potential but this is an oxidation; therefore, reverse the sign. The E value for the half reaction as written is +0.76 v. The other is a reducti...

**Chemistry need help**

I've checked every five minutes and no activity. So are you working on your homework?

**Chemistry need help**

Look up and post the reduction reaction and Eo for each half cell so we'll be using the same numbers.

**CHEMISTRY HELP !!!!**

The concentrations of the "ingredients" changes with time because one is being depleted and the other is being increased. E = Eo - (0.05916/n)log(1/M^x+) for each half cell.

**Chemistry**

I must be missing something. If you had 1.75g initially and 45.02% remains, then you have 1.75 x 0.4502 = ? g remaining after 85 days. Isn't that right?

**Chemistry**

k = .693/t1/2 t1/2 should be in days. Solve for k and substitute in the equation below. ln(No/N) = kt No = solve for this N = 45.02 k from above t = 85 days Solve for No which will be 100%.

**Chemistry**

There are 3.7E10 dps/Ci 1 Bq = 1 dps. So 18.9 uCi will be 18.9E-6 Ci. (3.7E10 dps/1 Ci) x 18.9E-6 Ci = ? dps Since 1 Bq = 1 dps, then ? dps = ? Bq Check out my thinking.

**CHEMISTRY HELP !!!!**

In #1, formation of Pb to Pb^+ should be smaller than Pb ==> PbSO4 because the formation of PbSO4 drives the reaction to the right. #2. dG = -nFEocell

**chemistry first year**

This is a limiting reagent (LR) problem. You can tell because amounts are give for both reactants. V is voltage, value, volume. Volume maybe. ........N2 + 3H2 ==> 2NH3 begin volume = 20 + 100 = 120 L. What volume NH3 will be produced if we use all of the N2 and and excess o...

**chemistry first yera**

HA + NaOH ==> NaA + H2O mols NaOH = M x L = ? mols HA = mols NaOH (look at the 1:1 coefficients) mols HA = grams/molar mass You know mols HA and grams, solve for molar mass.

**Chemistry**

[350g H2O x specific heat H2O x (70-25)]+[mass steam x (heat vap)] + [mass H2O from steam x specific heat H2O x (70-100)] = 0 Solve for x = mass steam The only unknown in that equation is mass steam and mass H2O from steam. I would let them = y and plug in the other numbers an...

**Chem**

If the density of the solution is 1.00 g/mL (and it may not be) it has a mass of 4200 grams. That x 0.031 = g Ag in the solution.

**Chemistry**

(P1V1/T1) = (P2V2/T2)

**chemtry 1**

Use PV = nRT

**chemistry**

You just need to set what you have equal to Kb for CN^- which isn't in any table but can be calculated this way. Kb for CN^- = (Kw/Ka for HCN) so (Kw/Ka for HCN) = (x)(x)/(0.8-x) and solve for x.

**TO ANY SCIENCE HELPER**

You know the half life for carbon 14 is 5,730 years. If you aren't told that in the problem you can look it up. You can calculate the rate constant (which you will need later) this way. k = 0.693/t1/2 k = 0.693/5730 = 1.21E-4 Substitute this k into the equation below. ln(N...

**Chemistry**

I don't think so. Pdry gas = 754mm - 18 mm = 736 mm and P gas in atm is 736/760 = ? Then PV = nRT or n = PV/RT = 736*0.0885/(760*0.08206*293) = ? Then mols = g/atomic mass or atomic mass = g/mols. Atomic mass Mg is about 24 and I obtained more than twice that.

**chemistry**

Do you mean 0.890 m or 0.890 M?

**TO ANY SCIENCE HELPER**

Why don't you post a problem and explain what you don't understand about it.

**chemistry**

2Al2O3 ==> 4Al + 3O2 mols Al2O3 = 742.5 g/molar mass Al2O3. Now use the coefficients in the balanced equation to convert mols Al2O3 to mols Al. Finally, g Al = mols Al x atomic mass Al.

**Chemistry**

I believe it has 4 non-bonding electrons. The two (paired) in the 2s and the two (paired) in the 2p.

**Chemistry**

I would go wth b. The C-O bond length is 112.8 pm according to wikipedia. The covalent radius for C is 0.77 and that of O is 0.73 according to my home periodic table.

**Chemistry**

.....Ag2CO3 ==< 2Ag^+ + CO3^2- I.....solid......0.....0.0001 C.....solid......2x.......x E.....solid......2x...0.0001+x Subsitute the E line into the Ksp epxression and solve for x = solubility in M. Then grams = mols x molar mass

**Chemistry**

I don't see any correction to go from pure O2 to air (21% O2 in air). A minor point. Since T is given as 26.4 I would have added that to 273.2. But if 882.6 is the correct answer I don't know how you came that close when not using the 21%. Perhaps you just didn't s...

**Radioactivity**

a. The rate of decay = 26,880 dpm (decays/min) b. k = 0.693/t1/2 Solve for k and substitute into the below equation. ln(No/N) = kt No = 26880 N = ? k from above t = 265 days c. 1 Bq = 1 dps Your sample is 26,880 dpm so 26,880 decays/min x (1 min/60 sec) = about 448 dps. ln(No/...

**Radioactivity**

The half live of C14 is only about 6,000 years and this is not suitable to a rock sample of that age. There is a section is the link below that talks about which methods are best for various ages. http://en.wikipedia.org/wiki/Radiometric_dating 2. k = 0.693/t1/2 Look up t1/2 i...

**chemistry**

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

**Chemistry**

right.

**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...........

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