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DrBob22
Answers (33)
Yes! Graph the data
You didn't convert to moles correctly. The molar mass of O2 is 22.4 L/mole ONLY at STP and 25C is not STP. Use PV = nRT, solve for n, then n/1L = M and I get 0.00126M. (Note: IF you could use 22.4, you should have divided 0.0309/22.4 = 0.00138 but that
What's the problem in doing this? I should be easy enough. Tell me your trouble; perhaps I can figure out the mental block.
5% NaCl w/v means 5 g NaCl/100 mL soln. So for 250 mL soln you will need 5g x (250/100) = ?? You can look at it another way. (grams/mL)*100 = % (X/250)*100 = 5 Solve for X.
(atomic mass O/molar mass C8H8O)*100 = ??
KOH is a strong base in aqueous solution; therefore, the (KOH) = (OH%-). For pH = 11.53, use pH + pOH = pKw = 14 to solve for pOH, then from pOH = -log (OH^-), solve for OH^-. That will give you the molarity of the KOH you want for that pH. Then M =
delta Hrxn = (sum delta H products)-(sum delta H reactants). Look these values up in a set of tables (one probably is in your text), substitute and solve.
Note the correct spelling of celsius. 1a. The work you did is ok but you didn't answer the question. The answer is no, the volume will not be doubled by doubling the celsius T. 2b. You halved the volume but the temperature went up. Something wrong here. I
I believe amides hydrolyze to the acid and NH3 CH3CH2COOH + NH3
If necessary, you CAN calculate the pH at the neutral point. HOH ==> H^+ + OH^- Kw = (H^+)(OH^-) = 1 x 10^-14 (H^+) = (OH^-); therefore, (H^+)^2 = 1 x 10^-14 and (H^+) = 1 x 10^-7 M pH = -log(H^+) = 7.0
Use the Ka expression (or the Henderson-Hasselbalch expression). Ka = (H^+)(CH3COO^-)/(CH3COOH) You are given CH3COO^- and CH3COOH, solve for H^+, then convert to pH.
A Bunsen burner, if hooked to a gas supply, is a device for heating materials.
No, you're done, except for multiplying everything by 2 to get rid of the 1/2 in 5.5 So you have H2O = 5.5 x 2 = 11 C2O4 = 3 x 2 = 6 Fe^+3= 1 x 2 = 2 K^+ = 4 So you have K4Fe(C2O4)6*11H2O I don't know what it is but something is wrong here BECAUSE Fe is
Is the solubility of KCl then 46 grams/100 mL at 70 degrees C? (I don't know if that is the unit for you OR if you were just copying the unit I gave as an example.) If 46 g/100 mL, then g in 50 mL = 46 x (50 mL/100 mL) = 23 g/100 mL.
You can't draw a graph on these boards but if its a table, can you either describe it or list the columns and temperatures? You can't type information on this board in columns but you can do it this way, using periods for spaces T........mass KCl/100 mL
You must have a table in your text or in your lab manual that gives the solubility of some of the salts at various temperatures. . I assume this is an experiment you are conducting. KCl is rather soluble.
Divide all the numbers by the smallest number, in this case that is 0.2068. Then round to the nearest whole number. If you have one that will round to 1/2, then multiply all by 2 to get rid of the 1/2. If you still don't get it, post your values for the
I have typed a response twice but the board won't let me post it. I don't know the trouble. I'll try this last time. When I say my tutors I mean those tutors who post on this board, and other boards. I read almost all of them and I learn something new
Answered below.
Take a 100 g sample; therefore, we will have the following: 20.51 g H2O. 52.60 g C2O4. 11.55 g Fe^+3.Note: you typed 98.15 but apparently divided by 98.18). 15.43 g K^+. Note: you typed 98.15, it should be 98.15--which gives the same number--BUT, I think
delta T = i*Kb*molality. You CAN calculate delta T for all four, then you will know quantitatively which is the lowest (actually you can't because concn is not given). BUT, you can do it the easy way. You won't know exactly what the freezing point is but
Generally speaking, write an extended summary, in which you focus on the conclusions you can draw from the experiment you performed. Some disciplines write an experimental, then discussion, then conclusions. I like to write the discussion and conclusion
answered elsewhere.
There isn't enough information to write the formulas for any of the compounds formed.QX and QZ is the best I can do; however, we don't know the valencies so we don't know how many of the Q adds to X or Z. We could write QaXb or QcZd where a,b,c, and d are
I would not expect a compound to be formed between Cr and element number 27 (Cobalt).
Both are isoelectronic with Ne; i.e., their outside shells are filled with eight electrons and all have the same number of electrons (that is the isoelectronic part).
Electron dot structures are difficult, if not impossible, to draw on these boards.
Yes, alkyl halides are made from alkanes (at least they are sometimes).
NaOH + HCl ==> NaCl + HOH strong base + strong acid. Determine mols NaOH = M x L = ?? Determine mols HCl = M x L = ?? Determine which reactant is in excess, then calculate molarity NaOH or HCl and pH from that. Post your work if you get stuck.
NaOH + HCl ==>NaCl + HOH Strong base + strong acid. calculate mols NaOH. M x L = ?? calculate mols HCl. M x L = ?? Determine mols NaCl and HOH produced. Determine excess NaOH OR HCl and pH from that. For NaOH + HC2H3O2 ==> NaC2H3O2 + HOH Strong base + weak
Zinc chloride is a white solid which is soluble in the aqueous solution present in HCl.
Ag2SO4 is silver sulfate LiF is lithium fluoride CS2 is carbon disulfide CaO is calcium oxide I2 is iodine
I would call chloroethane an alkyl chloride. If alkanes are defined as hydrocarbons, and this compound also has Cl, then it is not an alkane.
