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April 19, 2014

Search: NiSO4 + KOH precipitate

Number of results: 2,055

chemistry
A solution of .12 L of 0.160 M KOH is mixed with a solution of .3 L of 0.230 M NiSO4. the equation for this reaction is: 2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s) .89 grams of Ni(OH)2 precipitate form I need to know the concentration remaining in the solution of: 1...
Friday, October 14, 2011 at 5:44pm by Colin

Chemistry 101
A solution of .12 L of 0.160 M KOH is mixed with a solution of .3 L of 0.230 M NiSO4. the equation for this reaction is: 2KOH (aq)+NiSO4 (aq) ----->K2SO4 (aq)+Ni(OH)2 (s) .89 grams of Ni(OH)2 precipitate form I need to know the concentration remaining in the solution of: 1...
Friday, October 14, 2011 at 3:40pm by Colin

Chemistry
When we add a large amount of etilendiamine to a solution NiSO4 a lila precipitate is observed, what is this precipitate)
Friday, November 23, 2007 at 7:11am by Mary

Chemistry
A solution of 220mL of 0.200M Koh is mixed with a solution of 120mL of 0.240M NiSO4. 1.Determine limiting reactant? 2.Calculate the mass of precipitate formed? 3.Calculate the molarity of all ions left in the solution?
Sunday, October 9, 2011 at 5:10pm by anonymous

chemistry
A solution of 220mL of 0.200M Koh is mixed with a solution of 120mL of 0.240M NiSO4. 1.Determine limiting reactant? 2.Calculate the mass of precipitate formed? 3.Calculate the molarity of all ions left in the solution?
Sunday, October 9, 2011 at 6:01pm by anonymous

Chemistry--PLEASE HELP
A solution of 220mL of 0.200M Koh is mixed with a solution of 120mL of 0.240M NiSO4. 1.Determine limiting reactant? 2.Calculate the mass of precipitate formed? 3.Calculate the molarity of all ions left in the solution?
Sunday, October 9, 2011 at 7:06pm by anonymous

Chemistry
You wish to make a 0.200 M solution of NiSO4 (aq). How many grams of NiSO4 * 6H2O should you put in a 0.500 L volumetric flask? The answer is 26.3 g NiSO4 * 6H20 but I am not sure how to get there.
Tuesday, October 20, 2009 at 3:13pm by Kyle

College Chemistry
Is ta a typo for 5a? Is tghe solution is 5a(ta) NiSO4? mols NH3 = M x L = ? mols NiSO4 = (1/6) x mols NH3 M NiSO4 = mols NiSO4/L NiSO4. You know M and mols, solve for L.
Thursday, October 25, 2012 at 1:40am by DrBob222

chemistry
6% (is that w/w ?) means 6g NiSO4 in 100 g solution. That is 6g NiSO4/(6g NiSO4 + 94 g H2O). So 200 g sample would contain twice those amounts.
Tuesday, March 4, 2014 at 9:11pm by DrBob222

Chemistry
#1. mass water evolved = 2.08-1.22 = 0.86 grams. mass anhydrous NiSO4 = 1.22 grams moles H2O = 0.86/18 =0.0478 moles NiSO4 = 1.22/154.76 = 0.00788. Divide by the smaller to find the ratio to 1 mole NiSO4. 0.00788/0.00788 = 1 NiSO4 0.0478/0.00788 = 6.06 round these to 1 NiSO4 ...
Tuesday, October 6, 2009 at 11:52pm by DrBob222

College Chemistry
A solution of 110 of 0.150 koh is mixed with a solution of 290 of 0.250 niso4 . what is the precipate
Sunday, October 2, 2011 at 7:03pm by Anonymous

chemistry
HNO3 + KOH ==> KNO3 + H2O How many moles do you have? That's M x L HNO3. Look at the coefficients in the balanced equation. 1 mol HNO3 = 1 mol KOH; therefore, mols HNO3 = moles KOH. M KOH = moles KOH/L KOH. You know M KOH and mols KOH; solve for L KOH.
Monday, August 6, 2012 at 7:13pm by DrBob222

Chemistry
3M KOH x ?L KOH = 5M KOH x 0.5L KOH. Solve for ?L KOH.
Sunday, March 18, 2012 at 11:55pm by DrBob222

chem
you need to make .80 M solution of NiSO4 (aq). How many grams of NiSO4 * 6H2O should you put into a .50 L volumetric flask?
Monday, December 12, 2011 at 11:04pm by help

Chemistry
Formic acid is a monoprotic acid and KOH is a monohydroxy base; therefore, the reaction is 1:1 as follows: HCOOH + KOH ==> HCOOK + H2O So moles HCOOH initially = M x L = ?? moles KOH need to exactly neutralize that HCOOH is the same. Then M KOH = moles KOH/L KOH You know M ...
Monday, October 25, 2010 at 6:45pm by DrBob222

chemistry
1.00 times 10^2 WHAT of H2O? mols NiSO4 = grams/molar mass. molality = m NiSO4 = mols/kg solvent. Tnen delta T = i*Kf*m i = 2 for NiSO4. You know Kf and m, solve for delta T. Finally, subtract delta T from the normal freezing point to find the new freezing point.
Wednesday, January 9, 2013 at 12:36pm by DrBob222

Chemistry 22
Determine if the reaction produces a precipitate. For a precipitate mark a and if there is no precipitate mark b. 8. KOH(aq) + Fe(NO3)3(aq)  9. MgCl2(aq) + Na(NO3)2(aq)  10. NaBr(aq) + Ag(NO3) (aq)  11. Na3PO4(aq) + Ba(NO3) 2(aq) 
Saturday, December 4, 2010 at 11:39pm by Ivy

Chemistry
How many grams of NiSO4*6H2O are needed to prepare 200*10^2 mL of a 3.5*10^-2 M NiSO4 solution?
Tuesday, February 15, 2011 at 12:25pm by Mattie

chemistry
What mass of NiSO4 is present in 2.00 x 102ml of 6.00% NiSO4(aq)? (The density of the solution is 1.13g/ml.)
Monday, November 11, 2013 at 6:51pm by g

Chemistry
The answer is 18.4 WHAT? H2C2O4 + 2KOH ==> K2C2O4 + 2H2O mols H2C2O4 = 2.30E3 mols KOH needed = twice that. Look at the coefficients. 1 mol H2C2O4 = 2 mols KOH. M KOH = moles KOH/L KOH. Therefore, L KOH = moles KOH/M KOH. I get 18,400 L. I wonder if you made a typo in the ...
Thursday, June 14, 2012 at 7:06pm by DrBob222

chemistry
mols glutamic acid = 9.9/183.59 = estimated 0.05. mols KOH needed = estimated 0.05 MKOH = mols KOH/L KOH or L KOH = mols KOH/M KOH = estimated 0.05/0.96 = estimated 0.056 L or about 56 mL and that is one of the pKa values. Multiply that by 3 for total volume KOH required.
Friday, May 3, 2013 at 1:47am by DrBob222

chemistry
how many ML of each of the following solutions will provide 25.0g of KOH? A)2.50 M of KOH b)0.750 M of KOH c)5.60 M of KOH NEED HELP setting up problem!!
Wednesday, April 13, 2011 at 8:50pm by cat

Chemistry
KHC8H4O4 + KOH ==>K2C8H4O4 + H2O mols phthalate - grams/molar mass = ? mols KOH = same (look at the coefficients in the balanced equation.) M KOH = mols KOH/L KOH I get about 0.066 M. Post your work and I'll find the error.
Wednesday, October 3, 2012 at 11:30pm by DrBob222

Chemistry
KHP + KOH ==> H2O + K2P 1:1 ratio from the equation. mols KHP = grams/molar mass mols KOH = mols KHP M KOH = mols KOH/L KOH. Solve for L KOH.
Sunday, October 7, 2012 at 9:25pm by DrBob222

Chemistry
Calculate the molarity of the following solutions: 5.0g of KOH in 4.0L of KOH solution I went from grams of KOH to moles of KOH to molarity but still got the wrong answer... 5.0g KOH X 1 mol KOH/56.108g KOH = 5.0 mol KOH/56.108=0.0891 mol KOH/4.0L and got the answer of 0.02 M ...
Thursday, December 9, 2010 at 9:04pm by TT

Chemistry
CH3COOH + KOH ==> H2O + CH3COOK mols CH3COOH = M x L = ? mols KOH = the same (look at the coefficients in the balanced equation.) M KOH = mols KOH/L KOH.
Monday, April 8, 2013 at 7:07pm by DrBob222

Chemistry
Well... You know that KOH is a strong base... So it ionizes completely. KOH----> K+ +OH- So you need the molarity of KOH... So.. 1.100g/.5500 L x 1mol/56.11g KOH = 0.0356 M KOH So since KOH is strong... Concentration KOH, K+, and OH- = 0.0356M pOH=-log(0.0356) pOH= 1.45 pOH...
Monday, February 27, 2012 at 3:23pm by L.Bianchessi

Chemistry
You want how many moles KOH? That is M x L = ? moles KOH. moles KOH = grams KOH/molar mass KOH. YOu know molar mass and moles, solve for grams. A quick note to say that KOH is hygroscopic and it picks up water as well as CO2 from the air; therefore, one can't use KOH as a ...
Tuesday, November 8, 2011 at 10:34am by DrBob222

AP Biology
KOH doesn't get rid of the CO2 it simply combines with it to form a precipitate.
Sunday, November 11, 2007 at 4:10pm by Tu Madre

chemistry
KHP + KOH ==> K2P + H2O moles KHP = gams/molar mass. Using the coefficients in the balanced equation, convert mols KHP to mols KOH. M KOH = mols KOH/L KOH
Thursday, March 3, 2011 at 4:46pm by DrBob222

chemistry
A solution of 114 mL of 0.190 molecules KOH is mixed with a solution of 200 mL of 0.220 molecules NiSO4. a. write a balanced equation for the reaction that occurs. identify all phases in answer
Sunday, October 2, 2011 at 11:08pm by susie

Chemistry
A solution of 114 mL of 0.160 molecules KOH is mixed with a solution of 300 mL of 0.250 molecules NiSO4. a. write a balanced equation for the reaction that occurs. identify all phases in answer
Monday, October 3, 2011 at 1:25pm by Rheagan

chem
How many moles H2SO4 do you have. M x L = moles. Using the coefficients in the balanced equation, convert moles H2SO4 to moles KOH. Note that 1 mole H2SO4 = 2 mole KOH. Now M KOH = moles KOH/L KOH. Solve for L KOH.
Friday, March 11, 2011 at 8:53pm by DrBob222

AP Biology
The KOH absorbed the carbon dioxide and caused it to form a precipitate at the bottom of the vial.
Sunday, November 11, 2007 at 4:10pm by Ama

chemistry
Create a reaction: Na3PO4 + 3 KOH --> NaOH + K3PO4 You have 25 moles of Na3PO4.....you NEED KOH 25 moles Na3PO4 x 3 mol KOH/1 mol Na3PO4 = 75 mol KOH Convert from MOl to GRAMS 75 mol KOH x 56.106g KOH/ 1 mol KOH
Friday, May 6, 2011 at 11:54am by tc

chemistry
mols KOH = grams KOH/molar mass KOH. M KOH = mols/L soln.
Tuesday, January 22, 2013 at 6:57pm by DrBob222

Chemistry 11
calculate the mass of impure KOH needed to make up 1.20 L of 0.60 mol/L KOH(aq)? Assume the impure KOH is 84% KOH by mass and 16% water.
Sunday, June 16, 2013 at 5:13pm by LG

chemistry
moles KOH = 37g/molar mass KOH. moles OH- = moles KOH and M KOH = moles/L soln; therefore, pOH = -log(OH^-) and convert to pH.
Wednesday, March 9, 2011 at 12:24pm by DrBob222

Chemistry
H3PO4 + 3KOH ==> K3PO4 + 3H2O mols H3PO4 = M x L = 2.5 x 0.008 = ? mols KOH = 3x that. M KOH = moles KOH/L KOH.
Tuesday, July 24, 2012 at 12:54am by DrBob222

chemisty
How many moles HNO3 do you have? M x L = moles. Using the coefficients in the balanced equation, convert moles HNO3 to moles KOH. Then moles KOH = grams KOH/molar mass KOH; solve for grams KOH.
Wednesday, December 1, 2010 at 2:14pm by DrBob222

Chemistry
I worked this or a very similar problem last night for another post. I suspect you are the same author BECAUSE of the error in molarity. It makes no sense to write 0.250 molecules NiSO4. I'm sure you meant 0.250 molar NiSO4. Anyway, go back and look at your other post and you'...
Monday, October 3, 2011 at 1:25pm by DrBob222

chemistry
Which of the following aqueous solutions should NOT form a precipitate with aque- ous Ba(NO3)2? 1. K3PO4 2. K2SO4 3. K2CO3 4. KOH
Wednesday, September 19, 2012 at 5:12pm by cheri

Chemistry
KOH + HCl ==> KCl + H2O mols HCl = M x L = ? mols KOH = mols HCl (from the coefficients in the balanced equation) M KOH = mols KOH/L KOH.
Thursday, September 6, 2012 at 10:24pm by DrBob222

Chemistry
H2SO4 + 2KOH ==> 2H2O + K2SO4 mols H2SO4 = M x L = ? Look at the equation to convert mols H2SO4 to mols KOH. mols KOH = 2x mols H2SO4 MKOH = mols KOH/L KOH. Solve for M KOH.
Monday, October 8, 2012 at 10:37am by DrBob222

Chemistry
Has it occurred to you that if you had posted your work I could have found the error in half the time it takes to work the problem. H2SO4 + 2KOH ==> K2SO4 + 2H2O How many moles H2SO4 were used? That is M x L = moles H2SO4. Now convert that to moles KOH. From the equation, ...
Saturday, June 25, 2011 at 9:27pm by DrBob222

AP Chemistry
A chemist mixes 200.0mL of 0.800 M Fe(NO3)3 and 300.0 mL of 0.750 M K2Cr04 and enough of either 0.400 M KOH or 0.400 M Pb(no3)2 to end up with only two spectator ions in solution. a. How much precipitate will be formed b. How many mL of KOH or Pb(No3)2 must be used c. What are...
Sunday, January 16, 2011 at 12:21pm by qwerty123

Chemistry
Use the data in this table to calculate the solubility of each sparingly soluble substance in its respective solution. (a) silver bromide in 0.066 M NaBr(aq) mol L-1 I know how to do this one, found it to be 1.167E-11 which is correct. (b) nickel(II) hydroxide in 0.256 M ...
Friday, March 19, 2010 at 8:52pm by Val

chemistry
Find oxalic acid mass: m Given: (KOH hydrogen equivalence factor: 1) KOH Concentration: c = 0.1N = 0.1M KOH Volume: V = 0.250L (COOH)2 Molar Mass: w = 90.03517 0.00006 g/mol Stoichiometry: (COOH)2 + 2 KOH = (COOK)2 + 2 H2O .: m/w = cV/2
Sunday, September 8, 2013 at 3:11am by Graham

chemistry
HAc = CH3COOH = acetic acid KOH = potassium hydroxide mL KOH x M KOH = mL HAc x M HAc Substitute and solve for M KOH
Friday, November 11, 2011 at 9:25am by DrBob222

Chemistry 11
try this; n = Mv = 0.6x1.2 = 0.72moles so, 0.72moles is the total mole i.e. KOH and water. Since the impure KOH is 84%, then the mole for KOH is; 0.84 x 0.72 = 0.605moles so, m = nMr = 0.605x(molar mass of KOH).
Sunday, June 16, 2013 at 5:13pm by bonjo

chemistry
Find oxalic acid mass: m Given: (KOH hydrogen equivalence factor: 1) KOH Concentration: c = 0.1N = 0.1M KOH Volume: V = 0.250L (COOH)2 Molar Mass: w = 90.03517 0.00006 g/mol Stoichiometry: (COOH)2 + 2 KOH = (COOK)2 + 2 H2O .: m/w = cV/2 so: m = wcV/2
Sunday, September 8, 2013 at 5:01am by Graham

Chemistry
Which of the following aqueous solutions should form a precipitate with aqueous Fe(NO3)3? 1. KCl 2. KNO3 3. KOH 4. K2SO4
Tuesday, October 5, 2010 at 10:38am by John

Chemistry
Which of the following aqueous solutions should form a precipitate with aqueous Fe(NO3)3? 1. KOH 2. KCl 3. K2SO4 4. KNO3
Thursday, September 29, 2011 at 10:26pm by Yep

chemistry
Which of the following aqueous solutions should form a precipitate with aqueous Fe(NO3)3? 1. KNO3 2. KCl 3. K2SO4 4. KOH
Wednesday, September 19, 2012 at 5:11pm by cheri

Chemistry
KOH + HCl ==> H2O + KCl moles HCl = M x L = ?? Convert moles HCl to mols KOH using the coefficients in the balanced equation. Then MKOH = moles KOH/L KOH.
Tuesday, March 15, 2011 at 9:01pm by DrBob222

AP chemistry
which of the following aqueous solutions should not form a precipitate with aqueous ba(no3)? 1. KOH 2. K2CO3 3. K2SO4 4. K3PO4
Saturday, September 22, 2012 at 10:36am by Anonymous

Chemistry
Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). pKa1 = 1.3 and pKa2 = 6.7 a) before addition of any KOH b) after addition of 25.0 mL of KOH c) after addition of 50.0 mL of KOH d) after addition of 75.0 mL ...
Monday, March 11, 2013 at 9:26pm by KB

Chemistry
2KOH + H2SO4 ==> K2SO4 + 2H2O moles H2SO4 = M x L = ?? Convert moles H2SO4 to moles KOH. It will be ??mole H2SO4 x 2 = moles KOH. M KOH = moles KOH/L KOH
Monday, May 23, 2011 at 11:03am by DrBob222

Chemistry
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HClO(aq) with 0.200 M KOH(aq). The ionization constant for HClO can be found here. A. Before any addition of KOH B. After addition of 25.0 mL of KOH C. After 30.0 mL of KOH D. After 50.0 mL...
Sunday, February 10, 2013 at 5:35pm by Kathy

Chemistry
Calculate the pH for each of the following points in the titration of 50.0 mL of a 2.7 M H3PO3(aq) with 2.7 M KOH(aq). (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 50.0 mL of KOH (d) after addition of 75.0 mL of KOH (e) after ...
Sunday, November 17, 2013 at 7:49pm by Brody

chemistry
Take a liter of solution. 1.29 g/mL x 1000 mL = 1290 g. Part is water and part is KOH. The KOH is 30%; therefore, 1290 x 0.3 = ?? grams KOH. Determine mols KOH and that is mols/L and that is molarity.
Monday, September 10, 2007 at 2:35pm by DrBob222

intro to chem
Assuming HI + KOH = KI + H2O, one mole of KOH is needed to neutralize ach mole of HI 18.6ml of .197M HI contains .00366 moles of HI So, we need .00366 moles of KOH in 28.4mL of solution. .0284x = .00366 x = .129, so the KOH is .129M
Tuesday, August 13, 2013 at 12:12am by Steve

chemistry
...........in enough WHAT? water? moles KOH = 5.50g/molar mass KOH. (OH^-) = moles KOH/L soln pOH = -log(OH^-)
Monday, February 13, 2012 at 2:10am by DrBob222

chemistry
when 0.1M solutions of the following substances are mixed, which pair will precipitate? 1) Pb(CH3COO)2 and Mg(NO3)2 2) H2SO4 and KOH 3) ((NH4)2)CO3 and Na3PO4 4) NaOH and BaCl2 5) CaCl2 and Na2CO3
Monday, March 14, 2011 at 3:13pm by jim

Chemistry
An excess of aqueous AgNO3 reacts with 24.5 mL of 5 M K2CrO4(aq) to form a precipitate. What is the precipitate? Answer: Ag2CrO4 What mass of precipitate is formed? Answer in units of g.
Wednesday, March 23, 2011 at 1:47am by Mary

Chemistry
mols KOH = grams/molar mass M KOH = mols/L soln. M OH^- = M KOH pOH = -log(OH^-) and go from there.
Thursday, May 10, 2012 at 10:25pm by DrBob222

Chemistry
find the mole for HNO3 using n = MV where M is the molarity and V is the volume in Liters. then use the balance equation; KOH + HNO3 --> KNO3 + H2O the reaction is 1:1 so the mole you calculated for HNO3 is also the mole for KOH. i.e. mole KOH = mole HNO3. then convert the ...
Monday, June 10, 2013 at 1:29pm by bonjo

Chemistry
Find moles first: That is M x L = 0.20 x 0.500L = ? Then convert from moles to grams with moles/molar mass. You want to use the molar mass of NiSO4.6H2O: 0.10 mol x 262.84 g/mol = 26.28 g of NiSO4.6H2O
Tuesday, October 20, 2009 at 3:13pm by Aaron

AP Chemistry
How much do you want to make? about 1L? m = moles/kg solvent. m = 0.1 mol KOH/1000 g H2O grams KOH = 0.1 mol x molar mass KOH = ??
Wednesday, March 23, 2011 at 12:11am by DrBob222

Chem Help
Consider a solution that contains Ag+ , Ba2+, and Pb2+ each at a concentration of .2M a) You add NaCl until the concentration of Cl- is 5.0 x 10^-3 M. A white precipitate forms. How do you determine whether that precipitate was AgCl or PbCl2? b) If you seperated the ...
Sunday, April 6, 2014 at 8:42pm by Anonymous.

college chemistry
Consider the titration of a 50.0 mL sample of a 0.100 M solution of the triprotic weak acid citric acid (H3C6H5O7) with 0.100 M KOH. For citric acid, the three (3) acid dissociation constant values are ka1 = 7.40x10-3, ka2 = 1.70x10-5, and ka3 = 4.00x10-7, respectively. Given ...
Wednesday, April 21, 2010 at 5:59pm by Aubree

chem
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0*10^-8 a)before addition of any KOH b)after addition of 25.0 mL of KOH c)after addition of 40.0 mL of KOH d)after ...
Sunday, July 8, 2012 at 10:25pm by quinn

chemistry
You want pH = 11.70; therefore, pOH = 14.0-11.70 = ?? Then find OH^- from pOH and (OH^-) = ?? M. How many moles must you have if you want that molarity (OH%-). M x L = moles OH^-. What is the molarity of the 15.0% KOH solution? The mass of 1 L of solution is 1000 mL x 1.14 g/...
Sunday, February 22, 2009 at 11:23pm by DrBob222

chemistry
Technically the problem can't be solved because you have no units for 40.0 and 20.02. Assuming those are mL, the problem is solved as follows: 2KOH + H2SO4 ==> K2SO4 + 2H2O mols H2SO4 = M x L = ? mols KOH = 2x that (look at the equation coefficients) Then M KOH = mols KOH/L...
Tuesday, July 31, 2012 at 11:44pm by DrBob222

Chemistry
A 15% solution is 0.15 g KOH/g solution. And since the density is 1.14 g/mL, that is 0.15g/1.14 mL or about 0.13 g/1 mL. How much of that do you need? pH = 11.4. Convert to pOH and from there to (OH^-). I get close to 2.5 M. That is 2.5 mols KOH/liter of solution. How much do ...
Saturday, February 14, 2009 at 4:38pm by DrBob222

Chemistry
Write the equation and balance it. How many moles HNO3 do you have? That is M x L = ?? How many moles KOH are needed? From the equation you know it is the number as moles HNO3. Then moles KOH = grams KOH/molar mass KOH. Solve for grams.
Monday, April 18, 2011 at 11:00am by DrBob222

chemistry
All of these are done the same way. 1. Write and balance the equation. 2KOH + H2SO4 ==> K2SO4 2. Calculate moles. moles = M x L. moles H2SO4 = M x L = 0.825 x 0.0025 L = 0.00206 3. Using the coefficients in the balanced equation, convert moles H2SO4 to moles KOH. 0.00206 ...
Thursday, April 28, 2011 at 11:37am by DrBob222

chemistry
First off, I wrote NaOH instead of KOH. mL KOH = 26.54 M KOH = 0.0100 mL HCl = 25.00 M HCl = ??
Monday, June 21, 2010 at 7:39pm by DrBob222

chemistry
All of that is ok. And you are right about the dilution. The chemicals react with mols of one neutralizing mols of the other and the amount of water doesn't make any difference. My students always said, "But you've dilute the first chemical so it's not as concentrated." I ...
Monday, September 22, 2008 at 9:47pm by DrBob222

Chemistry
Use the data in this table to calculate the solubility of each sparingly soluble substance in its respective solution. (a) silver bromide in 0.066 M NaBr(aq) mol L-1 I know how to do this one, found it to be 1.167E-11 which is correct. (b) nickel(II) hydroxide in 0.256 M ...
Friday, March 19, 2010 at 10:07pm by Val

chemistry - Molar enthalpy
haha, yep, ok so q= 10,276.53 J and to get the answer in molar enthalpy should i divide q by the mole of KOH 10.0g KOH (1 mol KOH/ 56.105g) = 0.1782 mol KOH 10,276.53 J / 0.1782 mol KOH = 57,668.5J/mol = 57.6 kJ/mol Is this right?
Wednesday, February 29, 2012 at 9:58pm by Rose Bud

Chemistry 2
HNO3 + KOH ==> KNO3 + H2O a)HNO3 is 100% ionized; therefore, pH = -log(H^+) = -log(HNO3) = ? b) mols HNO3 = M x L = ? mols KOH added = ? mols HNO3-mols KOH = mols HNO3 remaining. M HNO3 remaining = mols HNO3/total volume. Don't forget total volume will be amount HNO3 you ...
Tuesday, April 2, 2013 at 10:44pm by DrBob222

Chemistry
A 25.00-mL sample of an H2SO4 (2 and 4 are subscripted) solution of unknown concentration is titrated with a .1328 M KOH solution. A volume of 38.33 mL of KOH was required to reach the endpoint. What is the concentration of the unknown H2SO4 (again, the 2 and 4 are subscripted...
Monday, January 4, 2010 at 1:22pm by coug

Chemistry
A 25.00-mL sample of an H2SO4 (2 and 4 are subscripted) solution of unknown concentration is titrated with a .1328 M KOH solution. A volume of 38.33 mL of KOH was required to reach the endpoint. What is the concentration of the unknown H2SO4 (again, the 2 and 4 are subscripted...
Monday, January 4, 2010 at 4:03pm by coug

Chemistry
You have three solutions, A, B, and C, each of which are believed to be one of the following: calcium hydroxide, potassium sulfate, and sodium chloride. Combining B and C results in the formation of a precipitate, but neither solution forms a precipitate with A. Also, adding ...
Sunday, September 29, 2013 at 11:15pm by Matt

chemistry
Explain how you would prepare these solutions from powdered reagents and whatever glassware you needed: a. 2.0 L of 1.5 M KOH the answer is: a. To prepare 2.0 L of 1.5 M KOH, measure out 168 g of KOH and place it into a 2 L volumetric flask. Add water to the mark. BUT HOW DID ...
Thursday, February 24, 2011 at 2:57am by lesa

Chemistry
This is a Lab Stoichiometry and Gravimetric Analaysis Balance equation: Na2CO3 + CaCl2 ---> CaCO3 + 2NaCl Precipitate(?): Not Sure Na2CO3 + CaCl2 Calculate the mass of the dry precipitate and the number of moles of precipitate produced ine the reactation? Empirical Formula...
Sunday, October 28, 2007 at 5:13pm by Monte;

Chemistry
balance the equation: 3KOH+ CrCl3>>3KCl + Cr(OH)3 notice for each mole of CrCl3, you use three moles of KOH. moles of CrCl3=.450*.675 moles of KOH=3*.450*.675 molarity of KOH= moles KOH/.550
Monday, September 27, 2010 at 7:28pm by bobpursley

chemistry
moles KOH = M x L = ? Convert moles KOH to moles H2SO4 using the coefficients in the balanced equation. ?mol KOH x (1 mole H2SO4/2 moles KOH) = ? mol KOH x (1/2) = x mol H2SO4 Now, M H2SO4 = moles H2SO4/L H2SO4. You know M H2SO4 and moles H2SO4, solvwe for L H2SO4 and convert ...
Monday, February 6, 2012 at 3:11pm by DrBob222

chemistry
OK this lab is on the formula of a precipitate and there were 8 test tubes of precipitate prepared obtaining different masses... but why does different masses of precipitate were obtained only for test tube IV and V?
Sunday, March 27, 2011 at 9:46pm by felaa

Chemistry
A solution contains the following ions: Hg2(2+) and Fe(2+). When potassium chloride is added to the solution, a precipitate forms. The precipitate is filtered off and potassium sulfate is added to the remaining solution, producing no precipitate. When potassium carbonate is ...
Friday, October 1, 2010 at 6:23pm by June

Chemistry
Write the equation and balance it. moles H2SO4 = M x L. Convert moles H2SO4 to moles KOH. Convert moles KOH to molarity. M = moles KOH/L KOH.
Wednesday, March 10, 2010 at 3:12pm by DrBob222

chemistry
Sodium fluoride and lead nitrate are both soluble in water. If 1.13 g of NaF and 0.232 g of Pb(NO3)2 are dissolved in sufficient water to make 1.00 L of solution, will a precipitate form? How do you know if there should be a precipitate or not? What is the chemical formula for...
Sunday, April 7, 2013 at 2:06pm by jeff

Chemistry
CH3COOH + KOH ==> CH3COOK + H2O moles KOH = M x L = ?? Convert moles KOH to moles CH3COOH. Since the coefficients are one in the equation, moles KOH = moles CH3COOH. Then M acid = moles acid/L acid.
Monday, May 23, 2011 at 5:51pm by DrBob222

chem
a.)How many moles of NaOH are required to prepare 2.00L od 0.380M solution? answer is 0.760 mole of NaOH (mole/l * 2.00L) b.)What will be the molarity of a solution if 3.50 g of KOH are dissolved in water to make 150.0 mL of solution? answer is 0.52 M KOH (gKOH/molar mass KOH=...
Friday, November 30, 2012 at 12:34am by Angel Locsin

Chemistry
I think your only problem is that you think Ni in NiSO4 is +1 but you know SO4 is -2; therefore, Ni must be +2 so Ni in Ni(OH)2 and Ni in NiSO4 are the same animal. (Ni^+2)(OH^-)^2 = Ksp I also think you can avoid a lot of confusion if you set up a chart; at least until you ...
Friday, March 19, 2010 at 8:52pm by DrBob222

chemistry
koh, kcl, no units on 0.5?? koh don't mean a think to me. AND you don't have punctuation except at the end. 0.5m KOH and 0.5m KCl(both in water) will have the same boiling point.
Thursday, December 2, 2010 at 11:11am by DrBob222

chemistry
H2SO4 + 2KOH ==> K2SO4 + 2H2O moles H2SO4 initially = M x L = ?? moles KOH (from the equation) = 2x that. M KOH = moles KOH/L soln. You have M and moles, solve for L and convert to mL.
Tuesday, February 22, 2011 at 6:52pm by DrBob222

chemistry
Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.8 M H3PO3(aq) with 1.8 M KOH(aq). before KOH after addition of 25, 50,75,100 ml KOH
Sunday, December 8, 2013 at 11:56am by Sofia

Chemistry
When sulfuric acid is added to magnesium nitrate, is the resulting product soluble or insoluble in water? Does a precipitate form? I think that the product is insoluble, and no precipitate forms. Is this right? The part I'm most unsure about is whether or not a precipitate forms.
Wednesday, December 3, 2008 at 3:31pm by Anonymous

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