Wednesday

March 4, 2015

March 4, 2015

Total # Posts: 3,004

**Chemistry(Please check)**

Oh Im sorry I missed that. CO3^2-^ is an answer listed. I was able to find that CN^-^ is a BL base but Im still not sure if it is a Lewis base. I do not think it is.
*March 24, 2012*

**Chemistry(Please check)**

1) Which statement below is correct (true)? a) Since HNO2 is a stronger acid than HF it must have a greater pKa. b) Since HOCℓ is a stronger acid than HCN, then OCℓ– is a stronger base than CN–. c) The dissociation (or ionization) of HOCℓ in water ...
*March 24, 2012*

**Chemistry(Please check)**

1) Which of the following is not amphiprotic? H20, NH4^+^, HSO4^-^, H2PO4^-^, all choices are amphiprotic. I think that all of these are amphiprotic. I know for sure that H20 and H2PO4^-^ are. 2) The ion HPO4^-^ has both a conjugate base as well as a conjugate acid. The ...
*March 24, 2012*

**Chemistry(Please just check, thank you)**

Ok thank you!
*March 24, 2012*

**Chemistry(Please just check, thank you)**

1) The approximate pH of a 3.0 X 10^-3 M solution of the strong acid H2SO4 is? I did pH= -log(3.0 X 10^-3) = 2.5 2)If HCLO2 is a stronger acid than HF, which is stronger than HOCL, then the order of strengths of the conjugate bases of these acids is? My answer is CLO2^-^ < ...
*March 24, 2012*

**Chemistry**

Write the chemical formula for the simplest heptane isomer and give its name, b) give the names of two heptane isomers with hexane as the base chain, c) give the names of names of the five heptane isomers with pentane as the base chain, d) give the name of the isomer with ...
*March 24, 2012*

**Chemistry**

What is the mass in kg of 20.0 L of diesel fuel? Assume the fuel has a density of 0.85 g/mL.
*March 24, 2012*

**Chemistry**

Thank you very much!
*March 21, 2012*

**Chemistry**

Consider the reaction: PCl3(g) + Cl2(g) PCl5(g). If [PCl3] = 0.78 M, [Cl2] = 0.44 M, and [PCl5] = 0.88 at equilibrium, what is the value of K? A.0.39 B.1.4 C.2.6 D.0.72 Thanks!
*March 21, 2012*

**Chemistry(Please check)**

ok
*March 20, 2012*

**Chemistry(Please check)**

For an experiment we had to test the equilibrium of BiCl3 - BiOCl First I had to mix a small amount of BiCl3 with 2mL of water. After doing this the color was cloudy white. Then I added6.0M of HCl drop-wise in which the color became clear. Then 25mL was added and the color ...
*March 20, 2012*

**statistics**

The average score on a science test was 72 with a standard deviation of 6. The average score on a history test was 89 with a standard deviation of 7. Compare the variability of the two classes.
*March 20, 2012*

**Chemistry(Please respond, thank you)**

I think thats why Im so confused because when added the HCl it only turned dark yellow and then our tube with the NaOH didnt even change. Ok so I see why it would shift to the left but im not sure what it was reacting with. From the equation the CrO4^2-^ is on the left side so...
*March 20, 2012*

**Chemistry(Please respond, thank you)**

oh ok I have to include in my answer if the process is exo or endothermic but I am not sure how I am suppose to know that. We then had to add 1 drop of 1.0M NaOh to a new test tube with 2.5mL of Na2CrO4 in it. I guess the color change was suppose to be orange even though my ...
*March 20, 2012*

**Chemistry(Please respond, thank you)**

Also the next question was which chromium ion is orange and which is yellow ans how do you know. So the dichromate is yellow and the Cr207^2-^ is yellow correct??
*March 20, 2012*

**Chemistry(Please respond, thank you)**

Ok so when the HCl is added it shifts to the right which is an exothermic process correct? It shifted to the right because the reactant is being used up??
*March 20, 2012*

**Chemistry(Please respond, thank you)**

For an experiment I had to add 1 drop of 1.0M HCl into a test tube that had 2.5mL of 0.10M Na2CrO4 and record the color. The original color of the Na2CrO4 was yellow. After adding the HCl the color became a very dark yellow. I know that the equilibrium shifted shifted to the ...
*March 20, 2012*

**Algebra 2**

Factor 12^2-3
*March 19, 2012*

**chemistry**

In the 1970s, when lead was widely used in "ethyl" gasoline, the blood level of the average American contained 0.25 ppm lead. The danger level of lead poisoning is 0.80 ppm. (a) What percent of the average person was lead? (b) How much lead would be in an average 80 ...
*March 19, 2012*

**Math(Please check)**

thank you for your help
*March 18, 2012*

**Math(Please check)**

2.5 X 10^(-3) = (.28)(.38) / x^2 Solve for x. 2.5 X 10^(-3) = 0.1064 / x^2 2.5 X 10^(-3) X 0.1064 = 266 sqrt 266 = 16.3 Did I do this correctly?? Thank you for your help.
*March 18, 2012*

**Math(Please someone respond, very confused)**

ok thank you!!!
*March 18, 2012*

**Math(Please someone respond, very confused)**

thank you I was looking for that post but couldnt find it.
*March 18, 2012*

**Math(Please someone respond, very confused)**

The original equation was 1.7e-3 = (0.0015 -x)^2 / (0.025 +1/2x)(0.025 + 1/2x) Square root of both sides is 0.0412 = (0.0015-x) / (0.025 + 1/2x) Solve for x and when I put in -0.001 or 1.0e-3 it says that it is incorrect, I dont know why??
*March 18, 2012*

**Math(Please respond, thank you)**

yes , it is 10^(-3). I sorry,im just so confused. I am typing it exactly as it is shown on my homework. They used().
*March 18, 2012*

**Math(Please respond, thank you)**

The equation is written as 1.7e-3 = (0.0015-x)^2 / (0.025 + 1/2x)(0.025 + 1/2x)
*March 18, 2012*

**Math(Please respond, thank you)**

The original equation was 1.7e-3 = (0.0015 -x)^2 / (0.025 +1/2x)(0.025 + 1/2x) Square root of both sides is 0.0412 = (0.0015-x) / (0.025 + 1/2x) Solve for x and when I put in -0.001 or 1.0e-3 it says that it is incorrect, I dont know why??
*March 18, 2012*

**Ecomonics(Urgent,please respond)**

The table below shows the market basket quantities and prices for the base year (year 1) Base year 1 Price in price Quantity base year yr 2 Product Pizza 15 $3 $3.75 t-shirts 4 $10 $9 rent 1 $500 $550 In year 1 the CPI was? In year 2 the CPI was? I know that cpi= (expenditures...
*March 18, 2012*

**Economics(Please respond, thank you)**

The table below shows the market basket quantities and prices for the base year (year 1) Base year 1 Price in price Quantity base year yr 2 Product Pizza 15 $3 $3.75 t-shirts 4 $10 $9 rent 1 $500 $550 In year 1 the CPI was? In year 2 the CPI was? I know that cpi= (expenditures...
*March 18, 2012*

**Chemistry**

Pure water boils at 100ºC (212ºF). When a substance is added to the water the boiling point increases due to what is referred to as the boiling point elevation. (a) What is the molecular mass of sucrose (C₁₂H₂₂O₁₁)? (b) If you add ...
*March 18, 2012*

**Math(Please help)**

The original equation was 1.7e-3 = (0.0015 -x)^2 / (0.025 +1/2x)(0.025 + 1/2x) Square root of both sides is 0.0412 = (0.0015-x) / (0.025 + 1/2x) Solve for x and when I put in -0.001 or 1.0e-3 it says that it is incorrect, I dont know why??
*March 18, 2012*

**Math(Please help)**

ok I tryed that to and for some reason it still says its wrong
*March 18, 2012*

**Math(Please help)**

yes its .0015 im sorry for the mistake, so it would be .0397=-40x +.5??
*March 18, 2012*

**Math(Please help)**

How did you get 39.5x?
*March 18, 2012*

**Math(Please help)**

For soem reason when I enter this answer on my online homework it says that its incorrect, I dont know why.
*March 18, 2012*

**Math(Please help)**

do for x I got 6.63e-4. Would you agree? Thank you for your help!!
*March 18, 2012*

**Math(Please help)**

oh ok and you got .0262 by subtracting 0.015 from 0.0412??
*March 18, 2012*

**Math(Please help)**

where did you get the 40x from??
*March 18, 2012*

**Math(Please help)**

0.0412 = 0.0015 - x / 0.025 + 1/2x Solve for x. 0.0412 = 0.015-x / 0.525x I am not sure where to go from here since there are two x's.
*March 18, 2012*

**Chemistry(Please respond, thank you!)**

Using this data, 2 NO(g) + Cl2(g) == 2 NOCl(g) Kc = 3.20 X 10-3 NO2(g) == NO(g) + ½ O2(g) Kc = 3.93 calculate a value for Kc for the reaction, NOCl (g) + ½ O2 (g) == NO2 (g) + ½ Cl2 (g) So I understand that 2NO + Cl2 + 1/2O2 = NO2 + 1/2 Cl2 but I do not ...
*March 15, 2012*

**Chemistry(Please respond)**

A study of the system, H2(g) + I2(g) == 2 HI(g), was carried out. Kc = 54.9 at 699.0 K (Kelvin) for this reaction. A system was charged with 2.50 moles of HI in a 5.00 liter vessel as the only component initially. The system was brought up to 699.0 K and allowed to reach ...
*March 15, 2012*

**Chemistry(Please respond, thank you!)**

ok so 6.66(473K)(0.08206) = 258.5 which would be 2.59 which is one of the answer choices. Does this seem correct??
*March 15, 2012*

**Chemistry(Please respond, thank you!)**

For the system CaO(s) + CO2(g) = CaCO3(s), I added 1.00 mol of CaO(s) to 1.00L of 0.500M CO2(g) at 200 oC. At equilibrium the [CO2] = 0.150M. What is the value of Kp for this reaction? Kp would = CaCO3/(CaO)(CO2) I do not know what to do with the 0.500M and 1.00mol. Chemistry...
*March 15, 2012*

**Chemistry - Please Check!**

Thank you.
*March 14, 2012*

**Chemistry - Please Check!**

mZnS = 0.018 x 97.47 g/mol mZnS = 1.71 g I am pretty positive the first part of is right but, if my final answer is wrong again, could you explain the question for me?
*March 14, 2012*

**Chemistry - Please Check!**

Determine the mass of zinc sulphide, ZnS, produced when 6.2g of zinc and 4.5g of sulphur are reacted. Equation given: Zn + S8 >> ZnS Balanced equation: 8Zn + S8 >> 8ZnS nZn= 6.2 g/ (65.39 g/mol) = 0.095 mol nS8= 4.5 g/(256.48 g/mol) = 0.018 mol nS8 = 0.095 mol Zn x...
*March 14, 2012*

**Chemistry(Please respond)**

Well Na ions were added so would that still shift to the left? How do you know if it's endo or exothermic I have to include that in my answer?
*March 14, 2012*

**Chemistry(Please respond)**

I did an experiment to study the effects of changes in conditions on an equilibrium system. The first system was the dissociation of acetic acid. HC2H3O2 -> H^+^ + C2H3O2^-^ <- We had to add 2mL of acetic acid into a test tube and we had to test the pH of the solution ...
*March 14, 2012*

**Chemistry(Please help)**

For the reaction, 2 SO2(g) + O2(g) == 2 SO3(g), at 450.0 K (Kelvin) the equilibrium constant, Kc, has a value of 4.62. A system was charged to give these initial concentrations, (SO3) = 0.254 M and (O2) = 0.00855 M, and (SO2) = 0.500 M. In which direction will it go? I think I...
*March 14, 2012*

**Chemistry(Please help)**

So I did 1/0.150 and got 6.66 so now I use this with RT. I am not sure what R is?
*March 14, 2012*

**Chemistry(Please help)**

For the system CaO(s) + CO2(g) = CaCO3(s), I added 1.00 mol of CaO(s) to 1.00L of 0.500M CO2(g) at 200 oC. At equilibrium the [CO2] = 0.150M. What is the value of Kp for this reaction? Kp would = CaCO3/(CaO)(CO2) I do not know what to do with the 0.500M and 1.00mol.
*March 14, 2012*

**Chemistry(Please help, thank you!)**

I am familiar with how to do the ICE table but I am not sure how to set one up for this question with the 2.5 moles.
*March 14, 2012*

**Chemistry(Please help, thank you!)**

A study of the system, H2(g) + I2(g) == 2 HI(g), was carried out. Kc = 54.9 at 699.0 K (Kelvin) for this reaction. A system was charged with 2.50 moles of HI in a 5.00 liter vessel as the only component initially. The system was brought up to 699.0 K and allowed to reach ...
*March 14, 2012*

**Chemistry(Please help, thank you!)**

So am I suppose to use the Kc values that are given??
*March 14, 2012*

**Chemistry(Please help, thank you!)**

Using this data, 2 NO(g) + Cl2(g) == 2 NOCl(g) Kc = 3.20 X 10-3 NO2(g) == NO(g) + ½ O2(g) Kc = 3.93 calculate a value for Kc for the reaction, NOCl (g) + ½ O2 (g) == NO2 (g) + ½ Cl2 (g) I do not understnad how I am suppose to use the Kc values given if ...
*March 14, 2012*

**Chemistry(Please check answer)**

I do not understand why do you 1/54.9??
*March 14, 2012*

**Chemistry(Please check answer)**

The equilibrium constant for the reaction, H2(g) + I2(g) == 2 HI(g) is 54.9 at 699.0 K (Kelvin). What is the equilibrium constant for 4 HI(g) == 2 H2(g) + 2 I2(g) under the same conditions? Note: the == indicates the equilibrium double arrow Since the second equation is ...
*March 14, 2012*

**Chemistry(Urgent, please respond, thanks!!)**

Thank you!!
*March 14, 2012*

**Chemistry(Urgent, please respond, thanks!!)**

1) For the reaction system, 2 SO2(g) + O2(g) = 2 SO3(g), Kc has a value of 4.62 at 450.0 K (Kelvin). A system, at equilibrium has the following concentrations: (SO3) = 0.254 M; (O2) = 0.00855 M. What is the equilibrium concentration of SO2? Is set this up as 4.62=(SO3)^2 / (...
*March 14, 2012*

**math**

i don't know. thats what i'm asking you
*March 13, 2012*

**Astronomy**

what are the largest stars in the HR diagram called?
*March 13, 2012*

**Chemistry(Can someone please answer!!)**

The decomposition of hydrogen peroxide in the presence of potassium iodide is believed to occur by the following mechanism: step 1 slow: H2O2 + I^- = H2O + OI^- step 2 fast: H2O2 + OI^- = H2O + O2 + I^- 1) What is the equation for the overall reaction? Use the smallest ...
*March 11, 2012*

**Math(Can someone please answer!!)**

Thank you!!!
*March 11, 2012*

**Math(Can someone please answer!!)**

ln [2.30E-2] = ln(0.102) - (k s-1)(2360 s) k = 6.31E-4 s -1 I did 2360k = 1n(0.102 * 0.023) 2360k = -6.055 then I divided 2360 by -6.055 but I didnt get 6.31e-4. Where did I mess up???
*March 11, 2012*

**Math(Can someone please answer!!)**

n [1.70E-2] = ln(0.144) - (k s-1)(3609 s) Solve for k. I have a TI30 calculator. Am I suppose to press just the ln button or do I have to press 2nd and then ln??
*March 11, 2012*

**Math**

ln [1.70E-2] = ln(0.144) - (k s-1)(3609 s) Solve for k. I have a TI30 calculator. Am I suppose to press just the ln button or do I have to press 2nd and then ln??
*March 11, 2012*

**Chemistry**

The decomposition of hydrogen peroxide in the presence of potassium iodide is believed to occur by the following mechanism: step 1 slow: H2O2 + I^- = H2O + OI^- step 2 fast: H2O2 + OI^- = H2O + O2 + I^- 1) What is the equation for the overall reaction? Use the smallest ...
*March 11, 2012*

**Chemistry**

The decomposition of hydrogen peroxide in the presence of potassium iodide is believed to occur by the following mechanism: step 1 slow: H2O2 + I^- = H2O + OI^- step 2 fast: H2O2 + OI^- = H2O + O2 + I^- 1) What is the equation for the overall reaction? Use the smallest ...
*March 10, 2012*

**Chemistry**

1/[N2O] = 1/(1.75M) + (1.10E-3M^-1 S^-1)(1852s) [N2O] = 0.383 M I know I posted a similar question but I am having trouble with this one as well. Should I do 1/1.75 first and then add 1.10e-3 and then multiply by 1852? Math(Thank you for your help) - john, Friday, March 9, ...
*March 9, 2012*

**Chemistry(Please help explain)**

The decomposition of hydrogen peroxide in the presence of potassium iodide is believed to occur by the following mechanism: step 1 slow: H2O2 + I^- = H2O + OI^- step 2 fast: H2O2 + OI^- = H2O + O2 + I^- 1) What is the equation for the overall reaction? Use the smallest integer...
*March 9, 2012*

**Chemistry(Urgent, please answer)**

Disregard question. I figured it out!
*March 9, 2012*

**Chemistry(Urgent, please answer)**

If you are given the slope and y intercept how do you find the rate constant of a chemical equation??? Would I multiply the slope and the y intercept???
*March 9, 2012*

**Math(Thank you for your help)**

So first I did multiplication (1.10e-3)(1852) and then divided 1/1.75 and then added and I cant get the answer.
*March 9, 2012*

**Math(Thank you for your help)**

1/[N2O] = 1/(1.75M) + (1.10E-3M^-1 S^-1)(1852s) [N2O] = 0.383 M I know I posted a similar question but I am having trouble with this one as well. Should I do 1/1.75 first and then add 1.10e-3 and then multiply by 1852?
*March 9, 2012*

**Chemistry(Thank you for your help!)**

O ok I didnt know it was that easy!!! Thank you!
*March 9, 2012*

**Chemistry(Thank you for your help!)**

The following info was obtained: [HI], M 0.535 0.268 0.134 6.70E-2 seconds 0 520 781 911 What is the average rate of disappearance of HI from t = 0 s to t = 520 s The answer is 5.13E-4 M s-1 but I do not understand how to get this!!!Could someone please explain!! Thank you!!
*March 9, 2012*

**Math(Please, please help)**

1/[N2O] = 1/(1.75M) + (1.10E-3M^-1 S^-1)(1852s) [N2O] = 0.383 M I know I posted a similar question but I am having trouble with this one as well. Should I do 1/1.75 first and then add 1.10e-3 and then multiply by 1852?
*March 9, 2012*

**Math(Please answer just a quick question!!!!)**

Ok how did you get 5.56e-3???
*March 8, 2012*

**Math(Please answer just a quick question!!!!)**

If the time elapsed is 4.26 min, what is the value of the rate constant, k? ln[R]t / [R]o = -0.0237 I am not sure how to slove this. Would I do ln 4.26 minutes times -0.0237???
*March 8, 2012*

**Math(Please help)**

ok thank you so much!!!
*March 8, 2012*

**Math(Please help)**

ln [5.01E-3] = ln(4.86E-2) - (1.80E-2s-1)(t s) Solve for t. t = 126 s First I did 4.86e-2 - 1.80e-2 and then on my calculator I pressed the 2nd button and then ln and got 1.031. Then I did the same thing to get the ln of 5.01e-3 and divided that by 1.031 but did not get 126 as...
*March 8, 2012*

**Math**

yes thats what it means and how I posted it is exactly how it is written on my homework but I just cant seem to get that answer.
*March 8, 2012*

**Math**

ln [5.01E-3] = ln(4.86E-2) - (1.80E-2s-1)(t s) Solve for t. t = 126 s First I did 4.86e-2 - 1.80e-2 and then on my calculator I pressed the 2nd button and then ln and got 1.031. Then I did the same thing to get the ln of 5.01e-3 and divided that by 1.031 but did not get 126 as...
*March 8, 2012*

**Math**

If the time elapsed is 4.26 min, what is the value of the rate constant, k? ln[R]t / [R]o = -0.0237 I am not sure how to slove this. Would I do ln 4.26 min X -0.0237?
*March 8, 2012*

**Chemistry(Please, please help)**

If the time elapsed is 4.26 min, what is the value of the rate constant, k? ln[R]t / [R]o = -0.0237 I am not sure how to slove this. Is Rt 4.26 min
*March 7, 2012*

**Chemistry (stoichiometry)**

18.5 grams of calcium chloride reacts with 25.7 grams of sodium phosphate and produces calcium phosphate and sodium chloride. What is the limiting reactant? (show work)
*March 7, 2012*

**Chemistry(Please help, thank you!)**

If the time elapsed is 4.26 min, what is the value of the rate constant, k? ln[R]t / [R]o = -0.0237 I am not sure how to slove this. Is Rt 4.26 min?
*March 6, 2012*

**Chemistry**

I need to calculate the valueof Kc for 5 test tubes for an experiment that I completed on determination of Kc. The equation used was Fe^3+ + SCN^-= FeSCN^2+ After creating an ice table for the first test tube my values for Fe and SCN were 6.5e-4M and for FeSCN I got 5.7e-5M. ...
*March 6, 2012*

**Chemistry(Please check)**

I need to calculate the valueof Kc for 5 test tubes for an experiment that I completed on determination of Kc. The equation used was Fe^3+ + SCN^-= FeSCN^2+ After creating an ice table for the first test tube my values for Fe and SCN were 6.5e-4M and for FeSCN I got 5.7e-5M. ...
*March 6, 2012*

**Chemistry(Please help)**

Nevermind I know what I did wrong. I had to divide my final answer by 1000 to get kJ.
*March 6, 2012*

**Chemistry(Please help)**

I did -1.23e4 / 8.314 = -1479.43 and I tryed 1.23e4 / 8.314 = 1479.43 but my online homework keeps saying that its incorrect.
*March 6, 2012*

**Chemistry(Please help)**

I had to make a graph using lnK vs 1/T adn find the slope. The slope was -1.23e4 and now I have to find Ea. Do I divide the slope by R
*March 6, 2012*

**Chemistry**

If the time elapsed is 4.26 min, what is the value of the rate constant, k? ln[R]t / [R]o = -0.0237 I am not sure how to slove this. Is Rt 4.26 min?
*March 5, 2012*

**Chemistry**

I had to make a graph using lnK vs 1/T adn find the slope. The slope was -1.23e4 and now I have to find Ea. Do I divide the slope by R?
*March 5, 2012*

**Chemistry(Please check)**

Right so for Fe I did (0.005M) (1.0mL Fe) / (7.0mL total of solutions in the tube) = 7.1e-4. So then my ICE table would say 7.10e-4 for both Fe and SCN because SCN for tube 1 had the same amounts and then I would subtract 7.1e-4 by the concentration of product which was 5.7e-5...
*March 5, 2012*

**Chemistry(Please check)**

Oh I see thats the total of SCN and Fe but we also used 0.1M HNO3 so for the first test time we used 5mL of HNO3 with the Fe, SCN so the total would be 7mL. Fe= (0.005M) (1.0mL) = x(7mL) = 0.000714 So 7.1 e-4 Correct. The teacher said to put the answer in 2 sig figs.
*March 5, 2012*

**Chemistry(Please check)**

Where did the 2mL come from?
*March 5, 2012*

**Chemistry(Please check)**

I think I know where I made a mistake. For the initial concentrations of reactants Fe and SCN it says that I can use the table that was provided to determine how much the total volume of each solution was used. The table that was provided provided the amount of Fe and SCN used...
*March 5, 2012*

**Chemistry(Please check)**

I completed a lab to find the determination of Kc. I have to find the concentrations of reactants at equilibrium using an ICE table. The equation that were are using is Fe^3+(aq) + SCN^-(aq) -> Fe(SCN)^2+(aq) I have to create 5 ICE tables because we used 5 different test ...
*March 5, 2012*

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