DrBob

NaCl is a solid. CH4 is a gas. The last one is a liquid. That should be enough of a hint.

15 years ago

Don't second guess yourself.

13 years ago

You should learn how to write arrows. Answer is D.

9 years ago

A is the best answer here. However, it isn't always true. Much depends upon how acids and bases are defined.

9 years ago

I don't like any of the answers but the best one here is A. NaCl dissolves in water because the NaCl crystal is surrounded by H2O molecules with the O end attracted to the Na ion and the H end attracted to the Cl ion. That attraction breaks the lattice en...

9 years ago

I can tell you that you do not make oxygen by heating MgO. You make MgO by heating Mg and O2. However, in the spirit of the question, I will answer as written. This is the Law of Conservation of Mass. You started with 10 g so you must end up with 10 g. .....

9 years ago

Acids have pH = 0 to pH = 7 so pH = -log(H^+). pH 0 then (H^+) = 1 x 10^0 pH 7 then (H^+) = 1 x 10^-7

9 years ago

density = mass/volume and (V1/T1) = (V2/T2) Reason it out.

9 years ago

pH = -log(H^+). Substitute and solve for (H^+). That will be the molarity in mols/L.

9 years ago

That's perfectly ok. You aren't plugging it in right. On my calculator, if you hit the 10^x key then -7, you get the answer of 1E-7. If you're going the other way (that is, given (H^+) = 1E-7, and you want the pH), then hit the log button, type in 1E-7 an...

9 years ago

You're Welcome :)

8 years ago

You have 25.0 mL x 1.00 M = 25 mmols or 0.025 mols HBr and KOH reacting. How much heat is generated? q = mcdT. m = mass = 25 + 25 = 50 grams. c = 1 cal/g. dT = delta T = 31.4-24.5 = 6.9 degrees.so q = 345 cal per 0.025 mols. How much is that for 1 mol? Re...

7 years ago

I answered this above. The answer of -14.8 kcal/mol is correct.

7 years ago

See post above with Tom.

7 years ago

That's a typo. It should be -13.8 kcal/mol. See above under post by Tom for my response.

7 years ago

tax = 77 x 0.063 = ? Total = 77 + ? = xx I agree with your answer.

7 years ago

yes

7 years ago

I can't give you a link but go to google and type in that word. You will gt plenty of information. Basically you take a triacid (such as stearic acid) and neutralize it with NaOH or KOH. The product is soap. That stuff is tough on the skin unless it's was...

7 years ago

Frankly, I don't like any of them.

7 years ago

I agree with the way the problem is worked above; however, I would round the 0.066 mols to 0.067 which makes the new density at 323 K to be 1.66 g/L which to to two significant figures is essentially the same. Here is much shorter way to work the problem....

7 years ago

Sorry, but I can't make sense of the question.

7 years ago

How many mols is 60 grams. That's 60/12 = 5. So -410 kJ/mol x 5 mol = ?

7 years ago

2Al + 3FeO ==> 3Fe + Al2O3 You have 13.2 mol FeO. Use the coefficients in the balanced equation to convert mols of what you want. 13.2 mol Al x (2 mol Al/3 mols FeO) = 13.2 x 2/3 = 8.8 and you answer is correct; however, I don't like the way you got there...

7 years ago

Do mean civil as in being civil to one another of civil as in construction (engineering).

7 years ago

Something is wrong. The heat of vaporization is given in units of cal/gram or cal/mol. .

7 years ago

OK. So the heat of vaporization (at whatever temperature that is)is 38.6 cal/gram CH2Cl2. That's why no T is given. How many grams do you have? That's 2 kg or 2,000 grams. So 38.6 ca/g x 2000 g = ? cal. And you thought this was complicated/

7 years ago

Nuclear particles stay together in the nucleus because some of the mass is convert to energy that binds them together. That is called the binding energy.

7 years ago

You need the equation for this.

7 years ago

This is stated in such a way it doesn't make sense to me.

7 years ago

If it is not an insulator it is called a conductor. . On Google search for insulators versus conductors

7 years ago

I don't know what the link is but dH rxn = (n*dHo products)-(n*dHo proucts) Then divide by 3 to get per mols C.

7 years ago

dHrxn = 973.49 kJ/mol x 4 = ? dHrxn = (n*dHof products) - (n*dHof reactants) You know dHrxn dHo of all but glycine. Substitute and solve for dHof glycine.

7 years ago

At the moment I am unable to post a link but if you will go to Google and type in rate law for i2 clock reaction, you should be able to go from there.

7 years ago

Animals breathe like humans.

7 years ago

Do you believe humans breathe in CO2 to convert it to carbohydrates?

7 years ago

If you think converting mass to energy means the particles do not weigh less when combined.

7 years ago

I would not have chosen that one. I think passage way refers to something like the breathing system we have.

7 years ago

How big is the statue. How much does it weigh or what are the dimensions. Wouldn't that make a difference?

7 years ago

Please check your problem. I believe there are typos in the numbers or you have omitted parts of the problem.

7 years ago

2H2O ==> 2H2 + O2 10.0 L O2 at STP = 10/22.4 mols = 0.446 mols or 0.446 x 32 = 14,3 grams O2. 96,485 coulombs will produce 32/4 = 8 g O2 so coulombs needed = 96,485 x (14.3/8) = ? Coulombs = amperes x seconds = 1.3*seconds. Plug in coulombs and A and solv...

7 years ago

Both of these answers is wrong. For #1, we have 2H2 + O2 ==> 2H2O. For all gas problems we can take a short cut and consider volume = moles. (The Bot worked it the long way, which is OK, but made an error in calculating the volume of H2 used in the reacti...

2 years ago

This is in error. 2 moles N2O4 will produce 1 mole O2. The bots error is that 1 mol N2O4 will produce 1/2 mol O2; therefore, the percent yield calculated by the bot will be 2 x 38.04 assuming all of the other calculations are correct.

2 years ago

However, note that ethanol and water form an azeotrope at 95.63% ethanol/ 4.37% water which means that 100% separation can NOT be realized using this technique alone.

2 years ago

Bot has an error somewhere. Here is how you do it. MCO3 + 2HCl ==> MCl2 + H2O + CO2 0.35 g + 0.1L x 0.1 M or 0.35 g + 0.01 mols HCl. How much extra HCl was in the solution unused to dissolve MCO3? That's HCl + NaOH = NaCl + H2O 0.01 - 0.15M x 0.02 L = 0.0...

2 years ago

The answer by Bot is true only if the volumes are additive. Technically they are not but for more practical purposes they may be considered to be additive.

2 years ago

x/2 = 8/9 cross multiply to obtain 9x = 16 then solve for x or x = 16/9 = 1.77777.........repeating

2 years ago

I should point out that this answer is true only for strong acids. It will not work for weak acids.

2 years ago

Bot, this is incorrect.

2 years ago

Yes, very funny. Here is the correct answer. One side is 9 and not 9.2 x^2 + 9^2 = 9.6^2 x^2 + 81 = 92.16 x^2 = 92.16 - 81 = 11.16 x = 3.34 mm I guess that's a small ladder.

2 years ago

Bot, you have failed to recognize that Q > Keq; therefore, the reaction proceeds from right to left. That is, if .............................Br2 --> 2Br I..........................0.063.....0.012, then Q = (Br)^2/(Br2) = (0.012)^2/(0.063) = 0.00229 > Keq...

2 years ago

Still a problem. First, the larger number of 0.0253 or 0.00775 can't be correct because the initial Br was only 0.012. The smaller number isn't correct either because 2 x 0.00775 = 0.0155 is larger than the initial amount of 0.012. Finally, note that your...

2 years ago

still not right.

2 years ago

You wrote: "At equilibrium, the concentrations will be: [Br2]eq = [Br2]0 - x [Br]eq = [Br]0 + 2x" No. At equilibrium (Br2)eq = (Br2)o + x (Br)eq = (Br)o -2x

2 years ago

James should do the math himself

2 years ago

Here is the problem in full.See my earlier post to show that the reaction will shift to the left in order to achieve equilibrium. ............................Br2 ==> 2Br I..........................0.063........0.012 C..............................+x.........

2 years ago

Hey Bot===a small math error. [A-]/[HA] = 10^(7.5 - 7.21) = 2.14 should be 1.95 but that won't change the volume much.

2 years ago

why are you adding? Shouldn't you be subtracting?

2 years ago

This is incorrect bot.

2 years ago

baloney. Bot, you need to use the H-H equation to solve properly.

2 years ago

shouldn't the ratio of A^-/HA = 1 after the reaction?

2 years ago

I believe 1.74 x 10^-5 is a better answer for 10^(-4.76) than 1.66 x 10^-5.

2 years ago

and the correct name is magnesium sulfate

2 years ago

Bot is right. The correct name is magnesium sulfate. Technically, epsom salts is the heptahydrate of MgSO4.7H2O

2 years ago

Of the choices listed the 2 is the best option, however, I usually refer to the valence of oxygen as 2-.

2 years ago

No, Bot is incorrect. The MOST volatile fraction contains the LEAST number of carbon atoms.

2 years ago

Bot, are you positive about that answer?

2 years ago

Right! If X has a valency of +a and Y as a valency of -b. then the formula of the compound will be XbYa.

2 years ago

Hey Bot. This is not the correct answer for x in Na2CO3.xH2O

2 years ago

still incorrect

2 years ago

Here is how to work the problem. Assume we take 1 mol Na2CO3.xH2O; thus we have 1 mol Na2CO3 and x moles H2O. mass Na2CO3.xH2O = approximately 268 from the problem. mass Na2CO3 = approximately 2(23) + 12 + 3(16) = approximately 106. Therefore, mass H2O =...

2 years ago

I assume that is 5.90 g MgCl2 produced. moles MgCl2 present in 5.90 g = grams MgCl2/molar mass MgCl2 = 5.90/95.2 = apprixunately 0.062. There are 0.062 mols MgCl2 x (6.022 x 10^23 molecules/1 mol) = approximately 3.7E22 molecles MgCl2. # Cl atoms = approx...

2 years ago

Bot, I believe there is a math error in the step below. I believe the correct ansewr is closer to 1.39 L V = (0.062 mol)(0.0821 L•atm/mol•K)(273 K) / 1 atm = 1.211 L

2 years ago

What does GPT3.5 and GPT4 mean

2 years ago

Incorrect. Yes, both AgNO3 and NH4Cl are soluble BUT not all chlorides are soluble; i.e. AgCl is not soluble;therefore AgCl will ppt and the reaction is AgNO3(aq) + NH4Cl(aq) ==> AgCl(s) + NH4NO3(aq) There is no reaction for the other one BECAUSE MgSO4 an...

2 years ago

Regarding the first question about water, 1 molecule of water + 1 molecule of water makes 2 molecules of water. There is no such thing as an atom of water. Regarding the equation, the solution by Bot does not balance the equation.

2 years ago

The equation still is not balanced. I think there is a typo in the initial equation. Here is the original. MnO4^-1 + SO3^2- + 12 H+ + 6 e^- → Mn4O^2- + SO4^2- + 4 H2O I think the author meant to write the following: MnO4^-1 + SO3^2- → MnO4^2- + SO4^2- und...

2 years ago

oxygen doesn't balance. charge doesn't balance. hydrogen doesn't balance. it isn't under basic conditions. the final equation may not have electrons.

2 years ago

still not balanced but closer. I'll do it right when I have time.

2 years ago

MnO4^-1 + SO3^2- → MnO4^2- + SO4^2- Divide into oxidation half and reduction half as follows: MnO4^- ==> MnO4^2- and SO3^2- ==> SO4^2- ================================== Balance each half MnO4^- + e ---> MnO4^= SO3^= + 2OH^- ----> SO4^= + 2e + H2O -------...

2 years ago

Bot, your equation is wrong. There is no need to add H2O to the MnO4^- --> MnO4^= My equation, reposted below, is correct. 2MnO4^- + SO3^= + 2OH^- -----> 2MnO4^= + SO4^= + H2O

2 years ago

does that make sense to you? -3.8 is closer to zero than -4.5

2 years ago

For the CH problem it the molar mass of the compound is listed as 78. Since CH is 12 + 1 = 13, then 78/13 = 6.0; therefore, the molecular formula is C6H6.

2 years ago

Bot, you forgot to multiply by 2 to obtain F^-. CaF2(s) ⇌ Ca2+(aq) + 2F-(aq) solid..........2.6E-5.......2*2.6E-5 Ksp = (Ca^2+)(F^-)^2 = (2.6E-5)(5.2E-5)^2 Ksp = 7.03E-14

2 years ago

Bot, did you make a math error?

2 years ago

I think you made another math error

2 years ago

I think your corrected answer still is incorrect. Bot, check out my response below. Do you agree or disagree. Starting with your equation of 6.3 x 10^-31 = 27x^4 x^4 = 6.3E-31/27 = 2.33E-32 x = (2.33E-32)^0.25 = 1.24E-8 M

2 years ago

Bot, I believe you made a math error in the quadratic formula. The equation you have of x^2 + 1.75 x 10^-5x - 4.375 x 10^-7 = 0 is correct but I have (OH^-) = approximately 0.00067 M which makes the pH = approximately 10.8. Check it out please.

2 years ago

Bot, you have an error in the quadratic formula. You are correct to this point. x = (-1.75 x 10^-5 ± √[(1.75 x 10^-5)^2 - 4(1)(-4.375 x 10^-7)]) / (2(1)). However, in the next step it appears you have multiplied by an extra 4.375 x 10^-7; i.e. in the quad...

2 years ago

Thank you.

2 years ago

please explain how there are 36 cards in the deck of 52 cards that are not numbered 2 through 9.

2 years ago

i don't get it. There are 32 cards from 2 through 9; therefore, there are 20 that are not numbered 2 through 9. 10, J, Q, K, A; i.e., 5 cards in each suit and 5 x 4 = 20 cards not numbered 2 through 9

2 years ago

Bot, I think your answer needs more work.

2 years ago

The question makes no sense to me. Soluble in what? water? The answers don't make sense to me either. ZnCO3 is not soluble in water but ZnSO4 is soluble. However, what is zinc sulfate acid? MgCO3 is not soluble in water. MgSO4 is soluble in water.

2 years ago

It's one times the 6

2 years ago

WOW. What a bunch of answers. The correct answer must be 1 cup sugar x (81 brownies/9 brownies) = 9 cups of sugar OR 9 brownies/1 cup sugar x ? cups sugar = 81 brownies ? cups sugar = 81/9 = 9 cups sugar OR (1 cup sugar/9 brownies) = (? cups sugar/81 brow...

2 years ago

The correct answer is 5/4 x 32 = 40 grams O2 needed.

2 years ago

Bot, the problem states that the solution is 1.33 m in methanol; therefore, there must be 1.33 moles methanol and not what you calculated.

2 years ago

No. A 1.33 molal solution means you have 1.33 moles methanol + 1000 g water and NOT 1000 g of SOLUTION. Therefore, there are 1.33 moles methanol + 55.5 mols water = 55.5 + 1.33 = 56.83 total moles. Xmethanol = mols methanol/total mols = 1.33/56.83 = appro...

2 years ago

Bot, your definition still is incorrect. The definition of a 1.33 molal methanol solution means that you have 1.33 moles methanol in 1000 g water for a total mass (which is not needed to work the problem). The part of the problem you have wrong is that "...

2 years ago

Great. Looks good to me.

2 years ago