Tuesday
July 7, 2015

Posts by DrBob222


Total # Posts: 49,412

AP Chemistry
whew! I'm glad I didn't goof.
April 20, 2015

AP Chemistry
I don't know. I don't think it can be PbCl2 forming since the two beakers are connected only with a salt bridge. The cathode surely must be the Cl2 since the Pt is an inert electrode and no conditions are given for the H2O there (that is I don' think we would worry...
April 20, 2015

AP Chemistry
If you know the anode and cathode already you can get the reaction from that. At the anode you have Pb ==> Pb^+ + 2e At the cathode you have Cl2 + 2e ==> 2Cl^- Add the two half equations for the cell reaction. You can add the phases.
April 20, 2015

san jac
V proportional to n V = 3.5 when n = 5 Therefore, if n = 2.5, V must be ...?
April 20, 2015

maths lit,life sciences,business studies,history
And what qualifications do you have?
April 20, 2015

Chemistry
Yes but I would keep the 6.29 since the 536 allows for three significant figures. Don't throw numbers away unless it's necessary.
April 20, 2015

chemisty
251J x (1 L*atm/101.325 J) = 2.48 L*atm work = -p(V2-V1) -(-2.48) = 2.7(V2-0.35) Solv for V2 in L.
April 20, 2015

Chemistry
You can reason this out. pH = pKa + log (base)/(acid) So if pH = pKa, then log (base)/(acid) must be zero. Set that as an equation to get log (base)/(acid) = 0 and 10^0 = ? ? must be 1; how do you get 1 out of log(base)/(acid). The only way to obtain 1 is for (base) = (acid) ...
April 20, 2015

Chemistry
Yes, 5.75E-8 is the answer for pH = 7.24. You should do pH = -log(H3O^+), then 7.24 = -log(H3O^+) -7.24 = log(H3O^+) (H3O^+) = 10^-7.24 = 5.75E-8
April 20, 2015

Chemistry
You calculated the pOH. The question asks for pH. After you have pOH, then pOH + pH = pKw = 14. You know pKw and pOH, solve for pH.
April 20, 2015

Chemistry
You took my response correctly and the correct answer is the HCN/CN pair because if Ka = 4.9E-10 then pKa = 9.31 which is quite close to the desired pH of 9.2; therefore, it is easy to make solutions in which the ratio of base/acid is close to 1. If you use the HH equation, ...
April 20, 2015

Chemistry
Actually you don't need to use the HH equation at all. The rule is that to make a buffer you want pKa to be within about 1 unit of the desired pH. So convert each of the Ka values to pKa values and choose the one that is within about 1 unit. And if you are good at scanning...
April 20, 2015

chemistry
mols ZnO desired = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols ZnO to mols O2. Then convert mols O2 to grams. g = mols x molar mass = ?
April 20, 2015

Chemistry
Please note the correct spelling of celsius. delta T = i*Kf*m The ice T is -0.4 so delta T is 0-(-.4) = 0.4 i for NaCl is 2 (two particles) Kf for water is 1.86 solve for molality.
April 20, 2015

Chemsitry
Use the Henderson-Hasselbalch equation.
April 20, 2015

gravimetric
[mass BaCrO4 ppt x (atomic mass Ba/molar mass BaCrO4)/mass sample)]*100 = %Ba For mass BaCrO4 substitute the following: (mg BaSO4 x 100/1000) into the top equation and solve for mass sample. Post your work if you get stuck.
April 20, 2015

gravimetric prob
AgCl is converted to Ag metal in the photodecomposition so the 10 mg that decomposed weighs 10mg x (atomic mass Ag/molar mass AgCl) = ? The other 10 mg does the same = ? So the new weight of the AgCl ppt is 704.7 mg - ? - ? = x mg AgCl + Ag.
April 20, 2015

gravimetric prob
Difference between initial and final weight perchlorate = mass H2O Difference between initial and final weight ascarite = mass CO2 %H = [(mass H2O x 2*atomic mass H/molar mass H2O)/mass original sample]*100 = ? % C = [(mass CO2 x atomic mass C/molar mass CO2)/mass original ...
April 20, 2015

Chemistry
NaCl added to ice lowers the freezing point of ice to below zero C, melting the ice takes energy, the energy must come form someplace; it comes from the water/ice mixture. The result is that the temperature of the ice/salt/water mixture is below zero. It can be MUCH lower than...
April 19, 2015

Chemistry
That looks ok to me but you're not allowed that many significant figures. If it were me I would round to 40.2 times.
April 19, 2015

chemistry
What do you think and why? Hint: Which will NOT change the concn of the substance in any way?
April 19, 2015

Science
No.
April 19, 2015

Molarity
16 g KNO3/molar mass KNO3 = mols KNO3. M of the diluted solution is mL1 x M1 = mL2 x M2 44.8 x 7.0 = 1100 x M2 Solve for M2. Then M = mols/L or L = mols/M. You know mols in the 16 g, you know M of the diluted solution, solve fo4 L of the diluted solution.
April 19, 2015

Chemistry
q = mass Cu x specific heat Cu x (Tfinal-Tinitial)
April 19, 2015

Chemistry 1002
I would do this. Calculate M of each of the solutions. 1.29 g/mL x 1000 mL x 0.38 x (1/98) = ?M 1.19 g/mL x 1000 mL x 0.26 x (1/98) = ?M Convert each to mols in the 715 mL. for the 1.29 it is ?M x 0.715 = ? for the 1.19 is is ?M x 0.715 = ? Now if you subtract the two mols you...
April 19, 2015

chem help
I assume you are to determine Ka from dGf values. Can you do that? What is Ka? For b. dG = -RTlnK. Plug in R, T, dG and solve for K. If you have the K values I can show you how to obtain % ionization.
April 19, 2015

chem
Which formula do you use? p = Kc*molarity Plug and chug.
April 19, 2015

Molarity
See your other post for the second one.
April 19, 2015

Molarity
mL1 x M1 = mL2 x M2 4.0L x 6.3M = 50L x M2 or you can do it with proportions by 6.3M x 4/50 = ?
April 19, 2015

Chemistry
dE = hc/wavelength. Plug and chug.
April 19, 2015

chem
(N2) = 4mols/2L = 2M (H2) = 2/2 = 1M (NH3) = 6/2 = 3M You should add all of the zeros on the numbers below. .......N2 + 3H2 ==> 2NH3 I.....2.0...1.0......3.0 C.....+x....+3x......-2x E.....2.0+x.1.0+3x..3.0-2x 3.0M-2x = 2.77M 3.0-2.77 = 2x and x = 0.115 Evaluate equilibrium...
April 19, 2015

chemistry
N2 + 3H2 ==> 2NH3 When gases are involved one may use a shortcut and use L as if they were mols. This is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants. I do these the long way. How much NH3 can be produced from 6 L N2 and all...
April 19, 2015

Chemistry
I will omit the phases. TiO2 + 4HCl ---> 2H2O + TiCl4 Since this is a limiting reagent problem (LR), I always do these the long way; i.e., calculate mols product formed from BOTH and take the smaller value. First 3 mol TiO2. 3 mol TiO2 x (1 mol TiCl4/1 mol TiO2) = 3 mol ...
April 19, 2015

chem 100
That doesn't help much. We need to know what to focus on to provide the best meaningful help. So look up the values of dHo and dSo. dHorxn = (n*dH formation products) - (n*dHo formation reactants. dSorxn = (n*dSo formation products) - (n*dSo formation reactants) dSsurr = -...
April 19, 2015

chem 100
Long question. How much do you already know how to do? Exactly what do you not understand. Regarding a part, what keeps you from looking up the values for dHo and dSo? Same for dGo for c.
April 19, 2015

chem
See your other post. http://www.jiskha.com/display.cgi?id=1429475277
April 19, 2015

Biochemistry
I assume the 4.77 has nothing to do with the problem or perhaps that was meant to be pKa for acetic acid. I've always used 4.74 for pKa acetic acid. You want the solution to be 0.2M so if you start with 0.2M HAc it will react with the KOH to produce acetate. Whatever you ...
April 19, 2015

Chemistry
I always have trouble with these because one never knows how negative numbers are counted when they increase or decrease. In addition the definition of lattice energy depends upon which book is used. Here is my reasoning. But first, the change is from 0.5 to 2 which is squared...
April 19, 2015

Chemistry
a. Br does follow 2N^2 b. There aren't enough electrons to fill it with more than 7. The first shell is full with 2, the second with 8 and the third with 18. That used up 28 and you have only 7 more electrons (35-28 = 7) so you place 7 electrons in the 4th shell. You don&#...
April 19, 2015

Chemistry
You know it can't be d. When is the last time you saw H2O decompose to H2 and O2 while you were watching. You must electrolyze H2O to make it do that which means you are adding energy to make it decompose so it isn't spontaneous. I might think increasing car prices are...
April 18, 2015

Science
Total mols HNO3 (too much) = M x L = ? - mols KOH added to correct = M x L = = mols HNO3 used to titrate the NaOH. Then M NaOH = mols NaOH/L NaOH. The answer is appox 0.5 M
April 18, 2015

Science
2H2 + O2 --> 2H2O When using gases, one may take a shortcut and use volume as if it were mols. 3.75L O2 x (2 mols H2/1 mol O2) = 3.75 x 2/1 = ? L H2.
April 18, 2015

Molarity question
You don't give enough information to calculate it but here is how you do it. mols FeCl3 = 0.825/162.2 = 0.00509 mols. How what was the volume you diluted this to? I will call it 250 mL but use what you did in the experiment. So for Fe in trial 1, you take 20 mL of the 250...
April 18, 2015

4th Grade
I would suggest you convert each of the answers into days. For example, 3 weeks x (7 days/1 week) = 21 days etc. For B, there are 24 hours in 1 day For C, there are about 30 days in a month For D, there are 24 hours in 1 day. But I assume you know all of those conversion factors.
April 17, 2015

Chemistry
Determine the empirical formula. Take 100 g sample which gives you 33.8 g C 1.42 g H 19.7 g N 45.1 g O ----------- Convert to mols. 33.8/12 = approx 2.8 C 1.42/1 = 1.42 H 19.7/14 = approx 1.41 N 45.1/16 = approx 2.8 O ------------- Find the ratio I think that is C2HNO2. The ...
April 17, 2015

type of reaction
Oh but it does. You're right, it is an acid reacting with a base to give a salt; however, the type is synthesis. Do you know the types? synthesis decomposition single replacement double replacement Here is a link. https://www.google.com/search?q=types+of+reaction&ie=utf-8&...
April 17, 2015

chemistry
..........2SO2 + O2 ==> 2SO3 I.........15.....10.......0 You don't give a Kp; I assume the reaction is to go to completion. SO2 can give 15 x 2 mols SO3/2 mols SO2 = 15 atm SO3. O2 can give 10 x 2 mols SO3/1 mol O2 = 20 mols SO3; therefore, SO2 is the limiting reagent ...
April 17, 2015

chemistry
Is this a multipart problem; i.e., is there another part of the problem in which Ea is calculated? or given? If so, what is the value of Ea in that part of the problem?
April 17, 2015

chemistry
For volume H2 unused at the end of the reaction. 4.0 mL Cl2 x (1 mol H2/1 mol Cl2) = 4.0 mL H2 used. Amount H2 left = 7.0-4.0 = 3.0 mL. So you will have 3.0 mL H2, zero mL Cl2, and 4.0 mL HCl when the reaction is finished.
April 17, 2015

Science
I agree with 1 and 3 but not 2 and 4.
April 17, 2015

Science
It appears to me that C is correct for #2. http://www.britannica.com/EBchecked/topic/40036/astronomical-unit-AU-or-au
April 17, 2015

Science
You make it so complicated to check your answers by placing them at the end. Place an asterisk by the answer so we don't need to scroll to the bottom for every question to see your answer.
April 17, 2015

Chemistry
No. I don't know what chart you are looking at but not all Pb salts are insoluble. PbI2, PbBr2, and PbCl2 are insoluble as are a number of other Pb salts. Pb(NO3)2, for example, is soluble.
April 17, 2015

biochemistry
You want how many mols? That's M x L = 0.2 M x 0.050 L = 0.01 mols. mols = grams/molar mass so grams HCl = mols HCl x molarmass HCl. That's approx 36.5 x 0.01 = about 0.365 g. Place that many grams HCl in a 50 mL volumetric flask, add DI water to the mark on the flask...
April 17, 2015

Chemistry
E = hc/wavelength You know h,c, wavelength, calculate E. Also f come either of two ways; 1. c = freq x wavelength or 2. E = h*freq.
April 17, 2015

Chemistry
See your post above. E = hc/wavelength
April 17, 2015

chemistry
part 2. I looked up the %H2SO4 in density of 1.425 g/mL and it is 52.63%. M H2SO4, then, is 1.425 g/mL x 1000 mL x 0.5263 x (1/98) = mols in 1 L and that is M.
April 17, 2015

oops--chemistry
Well, I screwed up big time. The problem says in big print H2SO4 and I used HCl.
April 17, 2015

chemistry
Equal volumes will average the % HCl so (30+70)/2 = 50% HCl. mols in 1000 mL will be 1.425 g/mL x 1000 mL x 0.50 x (1/36.5) = ? And that's the M since mols/L = M. Answer this Question
April 17, 2015

chemistry
By the way, I posted a response to your NaCl/CaCl2 ratio problem. If you haven't found it let me know and I can give youa link.
April 17, 2015

chemistry
2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2 mols Al = grams/atomic mass = 0.1 mols H2SO4 used = 0.1 mol Al x (3 mols H2SO4/2 mols Al) = 0.15 mol H2SO4 used. mols H2SO4 initially = 1.18 g/mL x 75 mL x 0.247 x (1/molar mass H2SO4) = approx 0.2 but you need a better answer than that. ...
April 17, 2015

chemistry
I don't understand the problem. Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2 There is the equation. I have no idea what "nature of the mixture" means but let's see what happens. How many grams Na2CO3 will the H2SO4 use? That's ml x N x milliequivalent weight ...
April 17, 2015

Science
Here is #2 first before we tackle #1. 2. How many types of molecules are represented? Is there a difference or what? How do I recognize an exact type? For example if there are two CO2, would that be considered 1 type? SO2 is one type, H2O is a second type, H2SO4 is a third and...
April 16, 2015

Science
An element is an element is an element. An element has ONLY one kind of atom and there are a few more than 110 elements known to man at this time. Here is a periodic table. Each symbol on the periodic table represents an element. That means that all of the atoms in that ...
April 16, 2015

Chemistry
PV = nRT and solve for n = number of mols. Then n = grams/molar mass. You know molar mass and n, solve for grams.
April 16, 2015

Science
This is all about Le Chatelier's Principle and it's easy if you just remember a couple of things. 1. The principle tells us that when a system at equilibrium is disturbed, it will react to undo what we did to it. a. An increase in pressure shifts it to the side with ...
April 16, 2015

science
Since dH is +, I include that in the equation this way. 2SO3 + heat ==> 2SO2 + O2 So an increase in heat causes the reaction to get rid of the heat and it fcan do that by shifting to the right. That's the way to use up the added heat. You're right. But if you cool ...
April 16, 2015

oops--Chemistry
I'll get in on the fray here. Irving has it wrong. Devron made a typo. I've copied Devron's work and replaced the typo part with a bold 5.0 M=[(0.0200L)*(3.0 M)+(0.0300L)*(5.0M)]/(0.0200L+0.0300L) The answer comes out to be about 4.2 which is the weighted average ...
April 16, 2015

To Jaimi--chemistry
I responded to your post(the one on ratio of +/- charges) that's way down the list (I think on page 2 or 3 by now). Here is a link. http://www.jiskha.com/display.cgi?id=1429203638
April 16, 2015

science
1. Since the values are given to more places that you have shown, I would recalculate and don't throw digits away. For example, pN2 you have only two places in your answer but more than that in the Ptotal and X. Same for H2 and NH3. Redo that. 2. Where do you go from here...
April 16, 2015

science
I should point out that the total doesn't add up to 100% (almost but not quite) so you may want to check your problem vs your post. %by volume = mole fraction or X. Therefore, XN2 = 0.96143; XH2 = 0.003506; XNH3 = 0.03506 pN2 = XN2*Ptotal pH2 = XH2*Ptotal pNH3 = XNH3*...
April 16, 2015

AP Chemistry
Did you use -p(Vfinal-Vinitial) 0.0100(5.1-3) = ? L*atm. The problem wants joules, convert by L*atm x 101.325 = ?J. Work should be a negative value; make sure to include the - sign when you enter into the database. Keep me posted.
April 16, 2015

AP Chemistry
Do you really care how many moles are there? 1 mol expanding or 100 mols expanding, the only thing that counts is volume difference and p.
April 16, 2015

chemistry
q = mass x Ccal x (Tf-Ti) q = 2.85 x Ccal x (29.19-24.05) Solve for Ccal. You will need to look up the value of q which is the heat of combustion for benzoic acid and substitute that for q. Note that heat of combustion is - but this heat is being added to the calorimeter so ...
April 16, 2015

Chmistry
Note that you have the equation going from right to left. If that's the way you want it, ok. This is Le Chatelier's Principle which, in basic words, tells us that when we stress a system at equilibrium it will try to undo what we did to it. I'm going to rewrite the...
April 16, 2015

chemistry
Is this an introductory course; how complicated can it get? I wuld think you could add enough NaOH solution (weak) to neutralize the acid and that makes it soluble. Then you can filter the sand and dump it. Back to the aqueous solution, add HCl to bring the benzoic acid back ...
April 16, 2015

chemistry
2C2H6 + 7O2 --> 4CO2 + 6H2O + 2834 kJ How many kJ does it take to do the water thing. q1 = heat needed to heat H2O from 54 C to 100. q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial) q2 = heat needed to boil away the water. q2 = mass H2O x heat vaporiation. Total heat ...
April 16, 2015

Science
You would do better to place asterisks next to your answers. Scrolling down to see your answer, back up to see the choices, back down to see answers is too time consuming.
April 16, 2015

Chemistry
It would be more than nice if you didn't switch screen names.
April 16, 2015

Chemistry
Reverse the Cd half cell and add to the Cu half cell as written. Calculate Eo cell. Then dG = -nFEcell. n is 2 and F is 96,485.
April 16, 2015

Chemistry
Ag.............0.8.........? Cu ...........+0.34.........0 H..............0..........-0.34 Ni...........-0.23..........? You just move everything down the page from Cu to h so you reduce everything by 0.34.
April 16, 2015

Chemistry
q1 = heat needed to raise T of solid ice from -10 to solid ice at 0 C. q1 = mass ice x specific heat ice x (Tfinal-Tinitial) q2 = heat needed to melt ice (change solid ice to liquid water). q2 = mass ice x heat fusion. Total = q1 + q2
April 16, 2015

Chemistry
If you have a Ag electrode dipping into the mixed solution and a Cu electrode dipping into the mixed solution, you get the battery reaction up front; i.e., Cu + 2Ag^+ ==> Cu^2+ + Ag(s)
April 16, 2015

chemistry
Obviously reduction is the gain of electrons.
April 16, 2015

chemistry
Oxidation is the loss of electrons.
April 16, 2015

chemistry
.........HX + NaOH ==> NaX + H2O mols NaOH taken = M x L = ? mols HX = mols NaOH since 1 mol HX = 1 mol NaOH in the balanced equation. Then mols HX = grams HX/molar mass HX. You know grams and mols, solve for molar mass.
April 16, 2015

Chemistry
mols CaCO3 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO. Now convert mols CaO to grams CaO with g = mols x molar mass = ?
April 16, 2015

chemistry
I don't see a question here. M = mols/L = 0.947/0.1 = ?M
April 16, 2015

chemistry
3.78 g NaCl/100 mL = 37.8 g/L solution and that's how many mols. 37.8/molar mass = approx 0.6 mols/L but you need answer than this estimate. This makes (Cl^-) = approx 0.6M You want to make 250 mL of BaCl2 and since BaCl2 has 2 Cl atoms/1 Bacl2 molecule, you only need 0.3 ...
April 16, 2015

chemistry
Let's assume for the moment the water is not there. How many mols do we want. That is M x L = mols = 0.5 x 60 = ? (I've used molarity since mols and equivalents for NaOH are the same). How many grams is that? grams = mols x molar mass IF IT WERE PURE STUFF BUT IT'S...
April 16, 2015

Inorganic chem
muriatic acid is HCl. HCl + NaHCO3 ==> NaCl + H2O + CO2 mols HCl = M x L = ? mols NaHCO3 = mols HCl since 1 mol HCl reacts with 1 mol NaHCO3. Then g NaHCO3 = mols NaHCO3 x molar mass NaHCO3.
April 16, 2015

chemistry
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. I do these the long way---easier to exlpain for me. AgNO3 + HCl ==> AgCl + HNO3 mols AgNO3 = grams/molar mass = ? mols HCl = M x L = ? Using the coefficients in the ...
April 16, 2015

Chemistry
Ba(NO3)2 + Na2SO4 ==> BaSO4 + 2NaNO3 mols Ba(NO3)2 = M x L = ? Using the coefficients in the balanced equation, convert mols Ba(NO3)2 to mols Na2SO4. Then M Na2SO4 = mols Na2SO4/L Na2SO4. You know mols and M, solve for L and convert to mL.
April 16, 2015

chemistry
I think what you want to do is this. Let's take a volume (choose anything but I'll pick 100 mL of the 0.2M NaCl) and calculate the amount of CaCl2 needed to meet the problem. 100 mL of 0.2M NaCl + xmL of 0.1M CaCl2, we have millimols + charge = 100*0.2 = 20 for Na in ...
April 16, 2015

Chemistry
Do you mean to boil all of the water away? That's q = mass H2O x heat vaporization Or do you mean to boil the first molecule of water? If so that takes much less heat.
April 16, 2015

Chemistry
All of these are worked with that magical formula of q = mc*delta T q = 90 x specific heat H2O x (Tfinal-Tinitial) specific heat H2O is 4.184 J/g*C Tf = 100 Ti = 0
April 16, 2015

Chemistry
It's the old q = m*heat vaporization m = 3.15g/18 = ? mols H2O heat vap = 44.0 kJ/mol q = ? kJ.
April 16, 2015

Science
It can't possibly be A. Why? There are 8 atoms Fe on the left and 5 on the right. There are 8 atoms S on the left and none on the right. Don't guess. Count the atoms on the left and see if that number is on the right. If not that one can't be the answer.
April 16, 2015

Science
I think you are right.
April 16, 2015

chemistry
That's my answer too but I think you should say the answer is 1.25%. By the way, it would be much better for everyone if you didn't change screen names. We can follow the thread better with the same name. It also helps on follow up questions to have the same screen name.
April 16, 2015

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