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March 30, 2017

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Posts by DrBob222

Total # Posts: 54,305

chemistry
We can't draw Lewis dot structure on this forum. I can try this way. Draw the Lewis dot structure for H2O. You have O with : on the left, top, bottom, right. An H goes on the left and the bottom which leaves the two electrons on the top and right undisturbed. Now a H ion (...
July 18, 2016

Chemistry
A is a limiting reagent (LR) problem. You know that when you are given amounts for BOTH reactants. You use BOTH to determine which is the limiting reagent, then use the one that is the LR. Right. You have 0.05 mol BaCl2 and you have 0.045 mols K2SO4. a. Convert mols BaCl2 to ...
July 18, 2016

Chemistry
mols Na2O = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Na2O to mols NaOH. That will be mols NaOH = 2 x mols Na2O Now convert mols NaOH to grams NaOH. grams = mols x molar mass
July 17, 2016

Chemistry
See you post above. Please use the same screen name. It help us help you.
July 17, 2016

Chemistry
See your other posts above. Please don't change screen names.
July 17, 2016

Chemistry
See your other posts. You get better help if you don't change screen names.
July 17, 2016

chemistry
You need to find the caps key and use it. Cu(OH)2 is insoluble.
July 17, 2016

Chemistry
5% m/v means 5 g solute in 100 mL solution. So the patient received 5 g x (5 mL/100 mL) = ?
July 17, 2016

Chemistry
0.4% m/v means 0.4 g solute/100 mL solution 0.4 g NaCl/100 mL x ?mL = 5 g NaCl Solve for ? mL
July 17, 2016

chemistry
1. 7% m/m means 7 g solute in 100 g solution. So you want 75 g solution. How many g solute will that be. That's 7 x (75/100) = 5.25 g solute. Place 5.25 g solute in 75-5.25 = 69.75 g H2O. I like to check these things to see that it really is 7%. (5.25/75)*100 = 7.0% You ...
July 17, 2016

chemistry
For KCl at 50 C. 42.6 g KCl x (35.0g H2O/100 g H2O) = 14.9 which means 14.9 g KCl can be dissolved in 35.0 g H2O. That means it will be saturated since 19.7 g KCl is greater than what will dissolve. 2 and 3 are done the same way.
July 17, 2016

chem II
Are these separate solution? NaCN + H2O ==> Na^+(aq) + CN^-(aq)
July 17, 2016

chemistry
Use PV = nRT and solve for n. Then n = grams/molar mass. You know grams and mols(n), solve for molar mass. Then molar mass/16 = x where Ox is the allotrope.
July 17, 2016

Chemistry
C2H5OH + 3O2 --> 2CO2 + 3H2O
July 17, 2016

typos--Chemistry
Note: Dr Rebel has reversed the Ka values (as well as using my estimates instead of the tabular values for Ka) AND calculated as if the acids were alone and not in a mixture.
July 16, 2016

Chemistry
Look up Ka values. Formic acid is about 10^-4 and HClO is about 10^-8 so formic acid is a much stronger acid. Also there is more of it. I would be tempted to ignore the HClO and calculate pH from HCOOH alone.
July 16, 2016

Chemistry
https://answers.yahoo.com/question/index?qid=20091202141141AAzWAa6 Answers are not in the same order but you get the full explanation here.
July 16, 2016

Chemistry
Do you have thoughts on this? It seems obvious to me that delta H is +. NH4NO3(s) + H2O ==> NH4NO3(aq) - heat
July 16, 2016

chemistry
We can't draw on this forum but here is a link that shows a diagram. https://www.google.com/search?q=activation+energy&ie=utf-8&oe=utf-8 The graph is a potential energy diagram. The X line shows the energy at the start, the Y line shows the energy at the end and the space ...
July 16, 2016

Chemistry HELP!!!
How many mols of each do you have? mols = grams/molar mass A is 4/4 = 1 mol B is 4/32 = 0.125 mol C is 4/2 = 2 mol D is 4/18 = 0.222 mol A mole of anything contains 6E23 particles. A is 1 x 6E23 = ? B is 0.125 x 6E23 = ? C is 2 x 6E23 = ? D is 0.222 x 6E23 = ? The one with the...
July 16, 2016

chemistry
When MgCl2 ionizes you get Mg^2+ and 2Cl^-. 2/3 of those ions are Cl^-. What's 2/3 of 6E23.
July 16, 2016

Chemistry
If it's 10 units long then one unit must be 625/10 = 62.5. Look up the four possibilities and see which has a molar mass of 62.5. For example, you know it can't be ethylene. H2C=CH2 is 28. Glucose is C6H12O6 and that is 180 so it can't be that. You can finish.
July 16, 2016

Chemistry
See my response below. Post any questions about my response there. Post your work if you run into trouble.
July 15, 2016

Chemistry
Maybe yes, maybe no. I can't tell exactly what you are asking so here is the problem is greater detail. Step 1 is the equation. Step 2 calculates the mols of what you have (you have 225 g SF6 to prepare) and you get that by mols = grams SF6/molar mass SF6 = 225/146 = 1.54 ...
July 15, 2016

Chemistry
1.54 mols for 225 g SF6 is correct. Go to the head of the class.
July 15, 2016

Chemistry
This a regular stoichiometry problem. Here are the steps; I suggest you print this out and keep it. This procedure will work any "simple" stoichiometry problem. It will work limiting reagent problems, also, but it needs two or three extra steps thrown in. 1. Writ and...
July 15, 2016

chemistry
mL1 x M1 = mL2 x M2 250 mL x 0.30M = mL2 x 0.10
July 15, 2016

chemisrty
e. I'll bet you could google, solubility PbO and find the answer quicker than waiting on someone to answer here..
July 15, 2016

Chemistry
How much O2 and N2 are you starting with?
July 15, 2016

Chemistry (DUE TONIGHT)
Thanks for all of the information but it would have helped if you had shown how you obtained the 1.27 number. I used this 97900 = 8.314*780*lnK and obtained approx 4E6.
July 15, 2016

chemistry practical
I don't understand the question.
July 15, 2016

chemistary
Please rephrase the question. It makes no sense as is. I would read that as 2 mol sulfur trioxide but that may not be what you have in mind.
July 15, 2016

chemistry
This is a limiting reagent (LR) problem. CaCO3 + 2HCl ==> CaCl2 + H2O + CO2 mols CaCO3 = grams/molar mass = ? mols HCl = grams/molar mass = ? Convert mols CaCO3 to mols CO2 Convert mols HCl to mols CO2 You will see that the mols CO2 don't agree; the correct answer in LR...
July 15, 2016

chemistry
"Structure" of bonding? The Cu to acetate part is essentially ionic, the bonds in the acetate are essentially covalent.
July 15, 2016

Science-light
If the arrangement is correct you really don't need a dark room; however, I assume the arrangement was not right. In this case, the light from the room would be measured and that isn't what you want.
July 14, 2016

college chemistry
How much do you want to prepare. I decide that first the I work these. You can decide later if you work it by molarity first. My way here. Let's say you want 1 L, then millimols = mL x M = 1000 mL x 0.2M = 20 mmols. pH = pK2 + log (base)/(acid) 5.7 = 7.2 + log b/a b/a...
July 14, 2016

@Steve--Chemistry
I don't think so. I answered the same way for Mariah about a week ago. The problem may be a faulty one OR she is not posting the numbers correctly.
July 14, 2016

Chemistry
Na2CO3 + 2HCl ==> 2NaCl + 2H2O + CO2 grams Na2CO3 actually titrated is 1.06 x (10/250) x 2(25/500) = ? mols Na2CO3 titrated = ?g/molar mass Na2CO3. mols HCl used = 2*mols Na2CO3 Then M = mols HCl/L HCl = ?
July 14, 2016

Chemistry
Its polar character.
July 14, 2016

Chemistry
They have different attractive abilities to the solvent used in the separation.
July 14, 2016

Chemistry
You're making it much too hard. How mcuh is rate increased when NO is doubled? That's rate = k[NO]^2[H2]. When NO is doubled rate is 4x; i.e., 2^2 = 4. When tripled rate is 9x; i.e., 3^2 = 9. What's the ratio of 9/4?
July 13, 2016

chemistry
CH4 + 2O2 ==> CO2 + 2H2O dHrxn = (n*dHo products) - (n*dHo reactants). Look up the dHo formation in your text, notes, net, and calculate.
July 13, 2016

Chemistry
Can you help. I don't know what the hubber process is? There is a Haber process for making ammonia but why would you test for ammonium. For that matter, what is ammonium?
July 13, 2016

Chemistry
Don't you follow up on your posts. Bob Pursley answered your first post hours ago.
July 13, 2016

Chemistry
How many mols do you need? That's mols = M x L = ? The mols = grams/molar mass. You know mols and molar mass, solve for grams.
July 13, 2016

Chemistry
a. Actually, no one knows; however, if all 4.80 dm^3 reacted AND the Cl2 gas was at STP, then it is 4.80/22.4 = ? b. Can't answer b,c,d. Is NaOH concd or dilute? Is it hot or cold? Here are a couple of reactions but there are others. 2 NaOH (conc., cold) + Cl2 = NaClO + ...
July 12, 2016

Chemistry
CuSO4 + Na2CO3 ==> CuCO3 + Na2SO4 mols CuSO4 = 1.5 mols CuCO3 = 1.5 x (1 mol CuCO3/1 mol CuSO4) = 1.5 grams CuCO3 = mols CuCO3 x molar mass CuCO3.
July 12, 2016

Chemistry
There is no reaction. There is a shift in the solubility equilibrium of Mg(OH)2 but no reaction.
July 11, 2016

Chemistry
pH = -log(H^+). pH 2.4; (H^+) = 3.98E-3 pH 4.4 = 3.98E-5 So it has been diluted by 100 times. 3.98E-3 x (10 mL/x mL) = 3.98E-5 Solve for X. You should have gone to 1000 mL. By the way, you should use the same screen name. We can help you better if you do.
July 11, 2016

Chemistry
This is a limiting reagent (LR) problem. I work these the long way but I think it's easier to explain this way. C + O2 ==> CO2 mols C = grams/molar mass = 14/12 = approx 1.1 but you need to redo this for a more accurate number. mols O2 = 58/32 = 1.8 Convert mols C to ...
July 11, 2016

Chemistry
I could just give you a formula that will do this; I memorized it years ago but here is a question by question approach. What is the definition of molarity. M = mols/L. So that's what we need to determine. How much does a L weigh? That's 1.114 g/mL x 1000 mL = 1114 ...
July 11, 2016

Chemistry
coulombs = amperes x seconds C = 3.00 x 48hr x (60 min/hr) x (60 s/min) = approx 518,000 C but you need to do that more accurately. We know that 96,485 coulombs will use 1 equivalent of Cu (1 eq of Cu is 63.54/2 = approx 32 g). So how much Cu is used? That's approx (63.54/...
July 11, 2016

Chemistry
This is done the same way as the Cu post I did for you above.
July 11, 2016

chemistry
What is your trouble with this one. EXACTLY what do you not understand. We HELP work problems but I don't see any work on your part.
July 11, 2016

chemistry
What do you think and why?
July 11, 2016

chemistry
Not only do you not get it, you apparently didn't even read how to do the first one.
July 11, 2016

chemistry
I'll be glad to help you through it BUT you must tell me what you don't understand about the problem and where you're having trouble. We HELP do homework.
July 11, 2016

chemistry
Isn't this just another LR problem? What are you having trouble with?
July 11, 2016

chemistry
Note that I showed how to do it in mols AND in grams.
July 11, 2016

chemistry
NaHCO3 + HCl ==> NaCl + H2O + CO2 mols NaHCO3 = grams/molar mass = 20/84 = approx 0.24. mols CO2 formed if NaHCO3 is the limiting reagent (LR)is 0.24 mols NaHCO3 x (1 mol CO2/1 mol NaHCO3) = 0.24 mols CO2. mols HCl = 22/35.5 = approx 0.62. mols CO2 formed if HCl is the LR ...
July 11, 2016

more dump---chemistry
Question 8 is the same kind as the first one you posted. Work it the same way. For the others, posting multiple questions with one posts USUALLY results in nothing being answered because it takes too much time for a single post. You might consider posting these as single posts...
July 11, 2016

Homework dump---chemistry
When magnesium metal and an aqueous solution of hydrochloric acid combine, they produce an aqueous solution of magnesium chloride and hydrogen gas. Using the equation, Mg (s) + 2HCl (aq) Imported Asset MgCl2 (aq) + H2 (g), if 72.9 g of Mg and 146.0 g of HCl are allowed to ...
July 11, 2016

@DrRebel----chemistry
No. Qc = 0.12; Kc = 0.8 Qc < Kc (or in your language Kc > Qc). Therefore, to make Qc = Kc requires products to increase and reactant to decrease. If you try shifting the reaction the other way you get a negative number for x and that can't be. Try working the problem...
July 10, 2016

chemistry
First you must determine which way the reaction will shift to react equilibrium. That is done with Qc. Qc = (PCl3)(Cl2)/(PCl5) = 0.12*0.12/0.12 = 0.12= Qc and Kc = 0.8 so Qc is too small which means products are too low and reactant is too high so shift must be from left to ...
July 10, 2016

chemistry
Do you mean with "final concentration" you want composition of NH3 or do you want the dissociation components; i.e., NH4^+ and OH^-. I will assume you want final NH3 composition and not the dissociation parts. mols #1 = M x L = ? mols #2 = M x L = ? final (NH3) = ...
July 10, 2016

Chemistry
You need to understand what's happening here. You add an excess of HCl to CaCO3 and it reacts completely but not all of the HCl is used. The NaOH is to determine how much of the HCl excess is there after all of the CaCO3 is gone. CaCO3 + 2HCl ==> CaCl2 + H2O + CO2 ...
July 10, 2016

CHEMISTRY
You had 7.46 grams to begin. You placed that in 1000 cc and pulled out 25 cc of that. So the amount in the 25 cc is all you titrated (you didn't titrate the entire 7.46g sample) so how much sample was in that 25 cc. That's 7.46 x 25/1000 =
July 9, 2016

CHEMISTRY
mass M2CO3 titrated = 7.46 x (25/1000) = approx 0.19 but you need a more accurate number for this and the all of the calculations that follow. M2CO3 + 2HCl ==> CO2 + H2O + 2MCl mols HCl = M x L = approx 0.0027 mols M2CO3 = 1/2*0.0027 = 0.00135 mols = grams/molar mass or ...
July 9, 2016

CHEMISTRY
H2C2O4 + 2NaOH ==> Na2CO2O4 + 2H2O 1.575 g H2C2O4.xH2O in 250 mL and 25 cc of that makes the sample titrated 1.575 x (25/250) = 0.1575grams. mols NaOH = M x L = approx 0.0025 but you need to redo all of this with better numbers. Convert mols NaOH to mols H2C2O4.xH2O. 1/2 * ...
July 9, 2016

science
You can read more about it here. https://en.wikipedia.org/wiki/Lifted_condensation_level height in meters = 125(T-Td) T is 28; Td is 18. Solve for h in meters. I get about 1250 feet.
July 8, 2016

Chemistry
Actually, the problem isn't stated properly. It should spell out if the answer needed is for mols H atoms or mols H2 molecules
July 8, 2016

Chemistry
Pressure in vessel 1 is p1 since the temperature hasn't changed. Pressure in vessel 2 is p2 = p1 x (t2/t1)
July 7, 2016

chemistry
You don't get it????? I responded to this yesterday with the quote that you were a better chemist than I if you can get NaOH from NH3 without doing something to it.
July 7, 2016

Chemistry
The problem is flawed. 18 g C can produce ONLY 66 g CO2.
July 6, 2016

chemistry
I don't understand the question, exactly. This may be what you are talking about. H2SO4 HOSO3H
July 6, 2016

Chemistry
q = mass Cu x specific heat Cu x delta T. Look up specific heat, substitute the numbers and solve for q.
July 6, 2016

Chemistry
The substance oxidized is the one that loses electrons. On the left C is -4 and on the right +4. On the left O2 is zero and on the right -2.
July 6, 2016

chemistry
Would you believe none? No H2 is involved in the equation. In fact the equation is wrong because a. it isn't balanced. b. the product and reactant are the same.
July 6, 2016

AP Chemistry
The three previous posts are all about electrochemistry. They are general with nothing to indicate what your problem is with these. Bob Pursley answered the first one. I'll be glad to help you through these but I'm not inclined to write a chapter or two, which really ...
July 6, 2016

AP Chemistry
From the previous work I'm sure you can see that it is important and why. I shall be happy to critique your thoughts.
July 5, 2016

AP Chemistry
S + O2 ==> SO2 Suppose you have 16 g S and 32 g O2 and they react. mols S = grams/atomic mass = 16/32 = 0.5 mol S. mols O = 32/32 = 1 mol O2. 0.5 mol S requires, from the coefficients in the balanced equation, 0.5 mol O2. Do you have that much O2. Yes, so S is the limiting ...
July 5, 2016

AP Chemistry
Because Mg combines with N2, also in the air, to form Mg3N2. So although most of the product would be MgO, it would be contaminated with some Mg3N2.
July 5, 2016

Chemistry
ASSUMING the acid you're using is 100%, then for 10% v/v you want 10 mL acid/100 mL solution. So scale that up from 100 mL to 2,000 mL. 10 mL x (2000 mL/100 mL) = ? Most acids are not 100%.
July 5, 2016

Chem
Coulombs = amperes x seconds. C = 2 x 16 x 60 = 1,920 For every (63.55/2) g Cu plated it requires 96,500 C. So (63.55/2) x (1920 C/96,500 C) = ?g Cu plated.
July 5, 2016

chemistry
I don't know how advanced this class is. If beginning your prof PROBABLY (I'm guessing) will want you to do this. H2SO4 ==> 2H^+ + SO4^2- So 0.01M H2SO4 gives 2 x 0.01 = 0.02M H^+ and you convert that to pH with pH = -log(H^+). However, it turns out that although ...
July 5, 2016

CHEMISTRY
I don't know how you obtained 998.4 mols. I might believe 0.9984. If you want the answer in tonnes, I would do this. A little unconventional but it saves a lot of work. If we work in tonnes, we never convert 64.0 tonnes to grams. Then we don't need to convert grams ...
July 5, 2016

chemistry
2KClO3 ==> 2KCl + 3O2 mols O2 needed = grams/molar mass = 48/32 = about 1.5? Using the coefficients in the balanced equation, convert mols O2 to mols KClO3. That's 1.5 mols O2 x (2 mols KClO3/3 mols O2) = 1.5 x 2/3 = about 1 So you need 1 mol KClO3 which is grams = mols...
July 5, 2016

chemistry
What about it?
July 5, 2016

Chemistry 1111
NaOH + KHP ==> NaKP + H2O mols KHP = grams/molar mass = ? mols KHP = mols NaOH (the ratio is 1 mol KHP to 1 mol NaOH in the equation). M NaOH = mols NaOH/L NaOH = ? Note: volume NaOH = 24.65 mL - 1.85 mL = ? mL. Convert to L.
July 5, 2016

chemistry
You have made a buffer by reacting NH3 and HCl. I assume "normal" conditions mean STP. mols HCl = 10/22.4 = approx 0.45 mols NH3 = 20/22.4 = approx 0.89 You need to redo all of these calculations. .......NH3 + HCl ==> NH4Cl I....0.89....0.45.....0 C....-0.45..-0....
July 5, 2016

chemistry
This is a hydrolysis equation. M NH4Cl = mols/L = grams/molar mass/L = approx 0.2 but you need to redo all of these calculations. ......NH4^+ + H2O ==> H3O^+ + NH32 I....0.2...............0.......0 C.....-x...............x.......x E.....0.2-x............x.......x Ka for NH4...
July 5, 2016

chemistry
See your other post.
July 5, 2016

chemistry
You're a better chemist than I if you can do that.
July 5, 2016

chemistry
HCl is a strong acid and the H^+ will be 5.00E-2 for the HCl. HCN is a weak acid. Calculate the H^+ contribution from that(decreased by the HCl as a common ion), then add the two together. Total H^+ = H^+ from HCl + H^+ from HCN. Convert to pH.
July 5, 2016

chemistry
This is done the same way you worked the NH3 gas and HCl gas problem above.
July 5, 2016

AP Chemistry
dG is negative. The reaction is spontaneous. dG = -; rxn spontaneous dG = 0; rxn about 50/50 dG = +; rxn not spontaneous in the direction shown but is spontaneous for the reverse rxn.
July 5, 2016

Chemistry
heat lost by metal + heat gained by ice = 0 [mass metal x sp.h. x (Tfinal-Tinitial)] + [mass ice x heat fusiion)] = 0 25*X*(0-150) + (9g*333.5 J/g) = 0 Solve for X
July 5, 2016

chemistry
A halogen is an element, as in F2, Cl2, Br2, I2. A halide is a salt of the halogen; i.e., a compound of a metal + halide. For example, Cl2, a halogen, combines with Na(a metal) to form NaCl (a halide--more specifically it is a sodium halide. In like manner you could have KCl, ...
July 5, 2016

chemistry
MgSO4 + H2O ==> Mg^2+(aq) + SO4^2-(aq)
July 4, 2016

chemstry
(huh)^2
July 4, 2016

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