# Posts by hannah

Total # Posts: 3,346

1) The Ka for acetic acid is 1.8e-5. What is the pH of a 3.18M solution of this acid? I did 1.8e-5 = x^2/3.18 x=sqrt 1.8e-5 X 3.18 = 7.56e-3 pH=-log(7.56e-3) = 2.12 The pH is 2.12 2) At 25 degrees celsius, the pH of a 1.75M NaCN solution (the Ka for HCN is 4.0e-10) is ? 4.0e-...

Life orientation
we will not just answer your homework questions. Give us some information to work with and then we can tell you if it is correct and add on to it

So the lowest pH would be the acidic solution correct?

Which 0.10M solution has the lowest pH? NaCl, KCN, LiF, RbClO, all have the same pH. I think that the answer is that they all have the same pH. Would you agree? Thank you!

Given the wage in dollar per day and the CPI for years 1 and 2 how do you calculate the nominal % increase in wage and the real % increase in wage? I am not sure what to do for this or what the formula is if any. Thank you for your help.

Oh ok I got it now thank you.

What is the [H3O+] for a neutral solution at 50 degrees celsius? I know that kw=[H3O+][OH-] but I am not sure what to do to get the H3O. The value I have for kw=5.76e-14 and I have the pH 6.62 but I do not know what the OH would be. Chemistry(Please help, thank you) - DrBob222...

I don't know if I put this in my calculator correctly but I got -0.82.

so I would take the - log of 6.62?

What is the [H3O+] for a neutral solution at 50 degrees celsius? I know that kw=[H3O+][OH-] but I am not sure what to do to get the H3O. The value I have for kw=5.76e-14 and I have the pH 6.62 but I do not know what the OH would be.

Given the wage in dollar per day and the CPI for years 1 and 2 how do you calculate the nominal % increase in wage and the real % increase in wage? I am not sure what to do for this or what the formula is if any. Thank you for your help.

Given the wage in dollar per day and the CPI for years 1 and 2 how do you calculate the nominal % increase in wage and the real % increase in wage? I am not sure what to do for this or what the formula is if any. Thank you for your help.

For an experiment on the effect of a buffer solution, I need to calculate the expected pH of the buffer. I know that I have to make an amounts table and use the Hasselbalch equation to find the pH. For the first part, 0.10M HCl was added to a buffer. The first addition was 5 ...

Organic Chemistry
I'm looking for a site with all the pentene isomers, and all the pentyne isomers. Anyone got any suggestions?

Chemistry
a) Write the chemical formula for the simplest pentyne isomer, b) give the name of the two isomers that have pentyne as the base chain, c) give the name of the one isomer that has butyne as the base chain, d) give the name of the 5 chain cyclic isomer, e) give the name of the ...

Chemistry
4 a) Write the chemical formula for the simplest pentene isomer, b) give the name of the two isomers that have pentene as the base chain, c) give the name of the three isomers that have butene as the base chain, d) give the name of the 5 chain cyclic isomer, e) give the name ...

Math
Mark wanted to measure the height of a nearby building. He placed a mirror on the pavement at point P, 80 feet from the base of the building. He then backed away until he saw an image on the top of the building in the mirror. If Mark is 6 feet tall, and he is standing 9 feet ...

Math
(4.44e-2 - x) (8.06e-2 - x) / (0.331 + 2x)^2 = 1.80e-2 I need to rearrange this equation so that I can get a,b, and c, for the quadratic equation but I am not sure how to do this. Thank you for the help. Math(Please answer, thank you) - bobpursley, Sunday, March 25, 2012 at 7:...

So I have to multiply (4.44e-2 -x )(8.06e-2 - x) by (0.331 + 2x)^2 or am I suppose to multiply 1.80e-2 by it?

(4.44e-2 - x) (8.06e-2 - x) / (0.331 + 2x)^2 = 1.80e-2 I need to rearrange this equation so that I can get a,b, and c, for the quadratic equation but I am not sure how to do this. Thank you for the help.

Chemistry
Determine the concentration in parts per million if 0.500 moles of NaCl are found in 35000g of a salt water solution. Okay, so I've done questions dealing with ppm and was able to work them out fine, but I can't seem to figure this one out. Any help you can give me ...

Ok so the only one im still not sure about is CO3^2-^. The -2 are lone pairs so I guess it is both Lewis and BL.

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.

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 ...

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 ...

Ok 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^-^ < ...

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 ...

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.

biology
C.

media
d. wire services

Chemistry
Thank you very much!

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!

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 ...

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.

I think that's why Im so confused because when added the HCl it only turned dark yellow and then our tube with the NaOH didn't 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 ...

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 ...

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??

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??

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 ...

Algebra 2
Factor 12^2-3

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 ...

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.

ok thank you!!!

thank you I was looking for that post but couldn't find it.

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 don't know why??

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().

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

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 don't know why??

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...

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...

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 ...

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 don't know why??

ok I tryed that to and for some reason it still says its wrong

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

How did you get 39.5x?

For soem reason when I enter this answer on my online homework it says that its incorrect, I don't know why.

do for x I got 6.63e-4. Would you agree? Thank you for your help!!

oh ok and you got .0262 by subtracting 0.015 from 0.0412??

where did you get the 40x from??

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.

political science/media
d. wire services

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 ...

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 ...

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??

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...

Thank you.

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?

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...

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?

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 ...

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...

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

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.

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.

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 ...

So am I suppose to use the Kc values that are given??

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 ...

I do not understand why do you 1/54.9??

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 ...

Thank you!!

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 / (...

math
i don't know. that's what i'm asking you

Astronomy
what are the largest stars in the HR diagram called?

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 ...

Thank you!!!

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 didn't get 6.31e-4. Where did I mess up???

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??

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??

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 ...

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 ...

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, ...

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...

Disregard question. I figured it out!