Wednesday

July 23, 2014

July 23, 2014

Total # Posts: 43,166

**Chemistry**

q = Ccal*(Tfinal-Tinitial) + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] Solve for Ccal Watch the units. If you use 4.184 J/g*C for H2O and you want Ccal in J/C, then you must substitute q in J, not kJ.

**chemistry**

(V1/T1) = (V2/T2) Remember T must be in kelvin. (210/200) = (360/T2) Solve for T2 in K

**physical chemistry**

X = 0.444*Y X = 2.96*12 Y = ?

**chemistry**

A lost of writing but no numbers. You can't calculate anything without numbers.

**physical chemistry**

mols Fe = grams/atomic mass = ?. There are 6.02E23 atoms in 1 mol so there will be ..... atoms in ? mol.

**physical chemistry**

Reason this out. mols = grams/atomic mass mols A = x/40 mols B = 2x/80 = x/40 Since there are the same number of mols of A and B there must be the same number of atoms of A and B. Therefore, since A contains y atoms then B must contain y atoms.

**physical chemistry**

mols C = 12g/12 = 1 1 mol contains 6.02E23 atoms. 98% of those will be C-12 and 2% will be C-14; therefore, number atoms C-14 = 6.02E23 x 0.02 = ?

**physical chemistry**

See your posts above.

**physical chemistry**

See your other posts.

**physical chemistry**

mols (NH4)3PO4 = 3.18 mols H x (1 mol (NH4)3PO4/12 mols H) = ? Then mols O is 4x that since there are 4 atoms O in 1 molecule (NH4)3PO4

**physical chemistry**

1 atm = 760 mm Hg = 101.325 kPa Make the conversion.

**physical chemistry**

mols CO2 = grams/molar mass There are 6.02E23 molecules in 1 mol CO2.

**physical chemistry**

no answer choices. abundance first isotope = 95.72% abundance other isotope is 100-95.72 = 4.28% Let x = mass other isotope (114.9041*0.9572) + (x*0.0428) = 114.82 Solve for x

**physical chemistry**

If 99% is 20 the the total must be approximately 20. Note: I stated at the top and worked down before realizing you had writer's cramp from all of your questions. This is not a place to dump your homework but for today I have helped with each one. In the future you need to...

**chemistry**

If plot ln k vs 1/T then slope = -Ea/RT

**chemistry**

Use the Arrhenius equation. Substitute k1 and k2 rates at the appropriate T and solve for Ea (activation energy); then use the same equation with Ea you found to solve for the new k.

**chemistry**

(1/A) - (1/Ao) = akt Let Ao = 100; A = 5; a = 1

**Chemistry - Electroplating - URGENT**

I believe you should have divided 52 by 3; the electroplating probably goes from Cr^3+ to Cr which is a change of 3 electrons. Then coulombs needed = 96,485 x (200/17.3) = ? C = amps x seconds but you have two unknowns. You don't know amps (usually given in the problem) or...

**Chem URGENT**

By the way you will get faster answers if you don't change screen names.

**Chem URGENT**

Do this in steps. 1. heat released in lowering steam from 114.5 to steam at 100 c. q1 = mass steam x specific heat steam x (Tfinal-Tinitial) q2 = heat released in changing steam at 100 C to liquid H2O at 100 C. q2 = mass x heat vap q3 = heat released in moving liquid H2O at 10...

**CHemistry**

0.5M x (3 SO4^2- ions/1 Al2SO4)3 molecule) = 0.5 x 3/1 = ?

**chemistry**

mols = grams/molar mass 0.007g/14 = ? mols C14 atoms There are 7 neutrons/atom; therefore, # mols x 7 = ? mols neutrons. There are 6.02E23 neutrons for each mol of neutrons; therefore, mols neutrons x 6.02E23 neutrons/mol = # neutrons. mass of those neutrons = # neutrons x mas...

**chemistry**

1 mol U weighs 238g There are 6.02E23 atoms in 1 mol. 1 atom x (238g/6.02E23 atoms) = ?g/a atom

**chemistry**

#/sec x #sec = 6.02E23 10^10/sec x #sec = 6.02E23 Solve for #sec and convert to years.

**chemistry**

Something is askew here. If you have 79% of one isotope and 21% of the other, that adds to 100% so there is no room for any other isotope.

**chemistry**

See your post on water answered by steve.

**chemistry**

Your question makes no sense.

**chemistry**

mols = grams/molar mass There are 6.02E23 molecules in a mole.

**chemistry**

Look the atomic mass up on the periodic table and add them. The answer is about 100.

**chemistry**

mols CO2 = grams/molar mass There are 6.02E23 molecules in a mol. Subtract 1E10^21 to find number left.

**chemistry**

What's your problem here. Are you dumping your homework?

**chemistry**

I'll be glad to help if you will explain what you don't understand. "None of it" is not an acceptable answer.

**chemistry**

2NaN3 ==> 2Na + 3N2 mols NaN3 = 88.8g/molar mass NaN3 = approx 88.8/65 = about 1.4 but that's an estimate. mols N2 produced = 1.4 mols NaN3 x (3 mols N2/2 mols NaN3) = 1.4 x 3/2) = estimated 2. Substitute n (2) into PV = nRT at the conditions listed and solve for V in l...

**Chemistry**

I think you can take your answer to the bank.

**Chemistry**

CuSO4.5H2O is soluble in water. With time the CuSO4 dissolves and begins to diffuse throughout the beaker. Left undisturbed, even after a long time, the bottom of the beaker will be quite blue and the upper portions will be much less blue in color.

**chemistry**

There are far too many chemicals for me to remember the solubility of each of them. If you have a table or a graph that gives solubility I can help. Otherwise, look up the solubility KClO3 and correct to 100 g.

**chemistry**

%H2O = (mass H2O/mass sample)*100 = ?

**chemistry**

https://www.google.com/search?q=ketohexose&client=firefox-a&hs=hp1&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&tbo=u&source=univ&sa=X&ei=sDmeU-MezbawBLbBgLAH&ved=0CCYQsAQ&biw=1099&bih=679&dpr=0.9

**chemistry**

http://en.wikipedia.org/wiki/Ketohexose

**analytical chemistry**

0.2810g Fe(NH4)2(SO4)2 x (1000 mg/g) x [1 mol Fe/1 mol Fe(NH4)2(SO4)2] x (1 mol Fe(NH4)2(SO4)2/284g) x (55.85g Fe/1 mol Fe) = ? Or without all the mishmash it is 0.2810 x (atomic mass Fe/molar mass Fe(NH4)2(SO4)2) x 1000 = 0.2810 x (55.85/284) x 1000 = ?

**algebra**

Here is how you do this with algebra. Let x = 10s digit and y = units digit. So the number is 10x + y and you know that if you reverse the number it is 10y + x. The problem tells you that 10x + y = 10y+x+18 which simplifies to 9x-9y=18 The second equation is x + y = 8. Solve t...

**chemistry**

http://www.jiskha.com/display.cgi?id=1402851831

**chemistry**

I did this for you below.

**Chemistry**

Na2CO3(s) + 2HCl(aq) = H2O(l) + CO2(g) + 2NaCl(aq) Use PV = nRT and solve for n = mols CO2 evolved at the conditions listed. Remember if you use kPa then R is 8.314. Remember T must be in kelvin. Convert mols CO2 to mols Na2CO3 knowing that 1 mol CO2 requires 1 mol Na2CO3. Con...

**oops--typo--chemistry**

Replace mass O = 3*6= 18 with mass O = 3*16 = 48

**chemistry**

HNO3. mass HNO3 = 1+14+(3*16) = about 63. mass O = 3*6 = 48 %O = (mass O/mass HNO3)*100 = ?

**chemistry**

Please type in the other information. What is "this" buffer?

**Chemistry**

mol NaOH = M x L = ? mols acid = mols NaOH mols acid = grams/molar mass. You know molar mass and mols, solve for grams.

**Chemistry**

Here is slightly different format. First determine the N of the HCl. The molar mass HCl is about 36.5 g/mol. 1.18g/mL x 1000 mL x 0.37 x 1/36.5 = ? and that is approx 12N Then use the dilution formula of mL1 x N1 = mL2 x N2 mL1 x 12 = 200 x 0.5 Solve for mL1.

**MAth**

4+sqrt(4+x^2)=x sqrt(4+x^2) = x-4 square both sides 4+x^2 = x^2-8x+16 You can finish.

**CHEMISTRY**

I'm not sure I agree completely with what you posted but do you have a question here?

**Chemistry**

I don't have the solubility tables nor do I have them memorized. I suspect you have either a graph or a table that gives you some of that information. If you will post that perhaps we can help.

**chemistry**

pH = 3.45 3.45 = -log(H^+) (H^+) = approx 3.55E-4. ...........HA ==> H^+ + A^- I.........0.1.....0.....0 C..........-x.....x.....x E.........0.1-x...x.....x where x = 3.55E-4 Substitute that into Ka expression and solve for Ka.

**chemistry**

The equation is wrong. 24.0 what? 0.982 what? KClO2 is not right. Straighten this out, watch for typos, perhaps we can help.

**chemistry**

Use PV = nRT and substitute the SECOND set of numbers to fine the NEW number of mols.

**Chemistry**

Did you get a definitive answer to the SO2/O2/SO3 question?

**Chemistry**

HCl + NaOH ==> NaCl + H2O millimols NaOH = 15 x 0.1 = 1.5 mmols HCl = 25 x 0.1 = 2.5 Excess HCl = 2.5-1.5 = 1 (HCl) = (H^+) = 1 mmol/40 mL = ? Convert to pH.

**check answer**

I believe that's right; or 500*sin(35)

**Chemistry**

I don't get it. It appears to me that any number one chooses can go there. Without O2 along with SO2 there is no way to establish equilibrium for SO2 alone or for O2 alone.

**Chemistry**

2C4H10 + 13O2 ==> 8CO2 + 10H2O 24.643 initial mass -24.592 after using -------- 00.051g mass butane used. mols butane = grams/molar mass = estimated 0.00088 but you need to get a more accurate figures. Using the coefficients in the balanced equation, convert mols butane to ...

**chemistry 2**

dSrxn = (n*dSproducts) - (n*dSreactants)

**Chemistry**

mols Mg = g/atomic mass = 5.2/24.3 = about 0.21 but you need a more accurate figure. mols HCl = L x M = 0.4 x 1M = about 0.4 Convert mols Mg to mols H2 = about 2 x (1 mol H2/1 mol Mg) = about 0.21 mol H2 Convert mols HCl to mols H2. That's 0.4 x (1 mol H2/2 mols HCl) = abo...

**Chemistry**

First, there is no way for me to know if these titrations have reached the equivalence point since the concn is unknown; however, I suspect what you really want to know is the concn of the unknown material Ba(OH)2 + 2HCl ==> 2H2O + BaCl2 mols Ba(OH)2 = M x L = 0.0217 x 0.05...

**Chemistry**

Do we start from scratch or do we have KNO3 to begin with? To start from scratch, take 25 mL 1M HNO3 and add to 25 mL 1M KOH. Evaporate to near dryness, place in a drying oven and leave until dry. Pure dry KNO3 is the result.

**Chemistry**

You don't say but I assume A and B are gases. total mols = mols A + mols B X(A) = mol fraction A = mols A/total mols. XB = mols fraction B = mols B/total mols pA = XA*Ptotal pB = XB*Ptotal

**Chemistry**

I agree.

**Chemistry**

I did, too.

**Chemistry**

That's right.

**Chemistry**

nope. E is true.

**Math Personal Finance**

doubled in ?? years.

**Math Personal Finance**

2000 x 0.10/4 = interest at end of quarter 1. Then 2000 + interest = total and I'll call that a. Then a x 0.10/4 = interest earned in quarter 2. interest + a = total at end of quarter 2 and I'll call that b. Continue for 3rd and 4th quarters.

**Math Personal Finance**

It will earn 0.55 x 4000 = ? annually. ? x 2.5 years = total interest earned. 4000 + interest earned = balance at maturity.

**engineeering**

You don't have a question here. And you can make your post much simpler if you don't write the underscore marks. We know C656H1707O756N11S where the subscripts are.

**oops--Chemistry**

I made a typo in that cubic equation. 4x^3-6.68E-7x^2+1.67E-8x-1.04E-10 = 0

**Chemistry**

It's close but not right. 1.67E-7 = (2x)^2(x)/(0.025-2x)^2 is what you started with and that's correct. Your error was in expanding the denominator. 1.67E-7 = 4x^3/(6.25E-4 -0.1x +4x^2) The next step is to multiply 1.67E-7 x the denominator, then collect the terms in d...

**chemistry**

See your post above which Bob Pursley answered.

**chemistry**

I can tell you how to do it if you can tell me how you get a concn of anything less than zero. I can see (OH^-) = 1 x 10^-80 but that is still more than zero. Not by much I'll grant you, but still more than zero.

**chemisrty**

5 is the answer. 5.5 to 3.5 the solution has become more acid by pH of 2 and since this is a log term (pH = -log(H^+) that's more acidic by a factor of 100. So OH must have decreased by a factor of 100.

**chemistry**

I think you meant this for the equation. I followed with the next two lines showing acid, base, conj acid, conj base. I think if you type the equatiion right you will see the answer. Also you need to restructure the answers to get them right. CO(NH2)2 + HOH ==> CONH2NH3^+ +...

**chemisrty**

Yes. Very good.

**chemistry**

mols KOH = grams/molar mass Then (KOH) = mols KOH/L solution Then (OH^-) = (KOH) since KOH is a strong electrolyte and dissociates completely. I would then do pOH = -log(OH^-) and follow with pH + pOH = pKw = 14 and solve for pH.

**Chemistry/physics**

What's the trouble here? Change 12.65 eV to joules = KE and solve for v. m = approx 9.11E-31 kg but you should confirm that since my memory isn't all that good any more.

**science**

I think 0.400 m/dm3 meant 0.400 mols/dm3 = 0.400 M.

**chemistry**

I don't see a question here.

**chemistry**

How many mols do you want? That's M x L = 0.400 x 0.500 = ? Then mols = g/molar mass. You know molar mass and mols, solve for mols.

**chemistry**

I'll show you how to do a and I expect you can do the rest but I can help you through them if necessary. Call benzoic acid HBz. ..........HBz ==> H^+ + Bz^- I.........1.0M....0......0 C..........-x.....x......x E........1.0-x....x......x Then 6.3E-5 = (x)(x)/(1.0-x) You...

**Chemistry**

250 x 0.82 = estimated 205 g starting FeS2. Convert 205 g to mols = about 2 mol but you should confirm that. Then mols H2SO4 = 2 mols FeS2 x (8 mols SO2/2 mol FeS2) x 0.95 x (2 mol SO3/2 SO2) x 0.90 x (1 mol H2SO4/1 mol SO3) x 0.90= mols H2SO4. Convert to grams.

**Chemistry**

M H2S2 initially = 0.0125/0.5 = 0.025M ............2H2S ==> 2H2 + S2 I...........0.025.....0.....0 C............-2x......2x.....x E.........0.025-2x....2x.....x Substitute the E line into the Kc expression and solve for x.

**physical science**

You mean it isn't obvious that I think the answer is concentrated?

**physical science**

I think the word that goes there is "concentrated" solution; however, dilute could be right under the right circumstances as well as saturated and unsaturated. I think the question could have been stated better.

**Chemistry**

Benzoate ion hydrolyzes in water to form a solution that is not neutral. If we call the benzoate ion Bz^-, then .........Bz^- + HOH ==> HBz + OH^- I........1.0..............0.....0 C........-x..............x.....x E.......1-x..............x......x Kb for Bz^- = (Kw/Ka for H...

**chemistry**

mols HAsc = 5/molar mass = approx 0.03 but you need to do it more accurately. ........HAsc ==> H^+ + Asc^- I......0.03......0......0 C.......-x.......x.......x E......0.03-x....x......x Substitute the E line into Ka expression for HAsc and solve for x, then convert to pH.

**Chemistry**

An organic acid is R-COOH where R stands for any organic group. A is an ester B is an alcohol etc.

**chemistry**

http://en.wikipedia.org/wiki/Cis%E2%80%93trans_isomerism

**Chemistry**

See your post above.

**chemistry**

6.3E1 mL or 6.3 x 10^1 mL To write any number in sci notation. place the decimal after the first digit, then count the places from the decimal to the end of the number and write "x 10^n" or "x 10^-n" where n stands for the number of places you counted. It i...

**Science**

Just for clarity; yes for A and yes for B.

**Science**

yes.

**Science**

pH + pOH = pKw = 14. You know pH and pKw, solve for pOH. Then pOH = -log(OH^-)

**chemisty**

Remember the definitions. Oxidation is the loss of electrons. So Cu goes from +1 to zero and iron goes from zero to +2. Which loses electrons?

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