Doc48

Newest questions and responses by Doc48
  1. Chemistry

    1mole O2 : 2 mole MgO That's all the work needed.

    posted on March 21, 2019
  2. Chemistry

    6.25 mol

    posted on March 21, 2019
  3. Chemistry

    53.3g

    posted on March 21, 2019
  4. Chemistry

    12.6-g CaCO₃ = (12.6-g/100-g/mol) = 0.126-mole CaCO₃ 0.126-mol CaCO₃ => 0.126-mol Ca⁺² + 0.126-mol CO₃²⁻ No. Ca⁺² ions = (0.126-mol Ca⁺² ions) x (6.02 x 10²³ Ca⁺² ions/ mol Ca⁺² ions) = 7.59 x 10²² Ca⁺² ions in 12.6-g

    posted on March 21, 2019
  5. Chemistry

    Great post DrBob!

    posted on March 21, 2019
  6. Chemistry

    Ca + Cl₂ => CaCl₂ is a 1 to 1 reaction ratio => 5 moles Cl₂ would then require 5 moles Ca.

    posted on March 21, 2019
  7. Chemistry

    For 0.40mole water in 1.4mole mix => 0.40/1.4 = 0.286 mole fraction of mix Then Partial Pressure water = 0.286(1.2 atm) = 0.343 atm water vapor pressure x 760-mmHg/atm = 261-mmHg water vapor pressure.

    posted on March 20, 2019
  8. Chemistry

    You sure the pressure value is asking for total pressure given only the partial pressure of water vapor? If so, this problem has too many unknowns for the data given. I'm guessing the 1.2 atm is total pressure and the problem is asking for the vapor

    posted on March 20, 2019
  9. chemistry

    If the energy number is on the product side of the equation, it is exothermic as the one you've posted. If the energy number is on the reactant side of the equation, it is endothermic. => H₂O(l) + 44Kj => H₂O(g)

    posted on March 20, 2019
  10. Chemistry (AP)

    This is a diprotic weak acid with a pKa₁ of 2.89 => Ka₁ = 1.3 x 10⁻³ and a pKa₂ = 4.40 => Ka₂ = 4.0 x 10⁻⁵. With such a low Ka₂, it is assumed (generally) that all of the Hydronium ions come from the 1st Ionization Step. For 0.20-g

    posted on March 20, 2019
  11. chemistry

    If the gas vapor mix is 0.2 mol A and 0.5 mole B where the total pressure is 40-mmHg, The partial pressure of A will NOT be 20-mmHg. It's more like 11.428-mmHg and B will be 28.572-mmHg. 0.2 mole A + 0.5 mole B => 0.7 mol mix. mole fraction A = (0.2/0.7) =

    posted on March 20, 2019
  12. Chemistry

    Rate = k[N₂O₅]¹ a. Rate 1 = (6.08 x 10⁻⁴ s⁻¹)(0.200 mol-L⁻¹) = 1.36 x 10⁻⁴ mol/L-s b. You do ‘b’

    posted on March 20, 2019
  13. Chemistry

    Bromothymol Blue

    posted on March 20, 2019
  14. Chemistry

    The last entry above is fm Doc48

    posted on March 20, 2019
  15. chemistry

    That is correct, DrBob ... If ya need it in joules/gram => ΔHᵥ = 2300 j/g

    posted on March 17, 2019
  16. chemistry

    The short version is... From a physical perspective,, there are 3 principle forces (that we know of) in the universe that give rise to three types of particle-particle interactions (bonds) that hold matter together as we know it in the scientific

    posted on March 17, 2019
  17. chemistry

    Methyl Amine & water coordinate to form methylammonium hydroxide (H₃CNH₃OH), a weak base (Kₐ =2.3 x 10ˉ¹¹) that ionizes ~0.6% as a 0.10M solution in deionized water as follows… H₃CNH₂ + H₂O => H₃CNH₃OH < = > H₃CNH₃⁺ + OHˉ Formic

    posted on March 17, 2019
  18. Chemistry

    Right on! Dr Bob:-)

    posted on March 15, 2019
  19. Chemistry

    True, you should be aware of the audience, but given results & responses to questions should also be accurate and defendable within the forum, and in this case, they were not. This is what gave me pause and prompted my comments. No apology necessary. :-)

    posted on March 15, 2019
  20. Chemistry

    Also, on sig figs, for problems, rounding is should be related to 'DATA' given. That is, do not use conversion factors or universal constants. DATA reflects the instrumental & experimental measurements used in the data collection process. For your problem

    posted on March 15, 2019
  21. Chemistry

    One of the best 10 minute videos on significant figures I've seen is on YouTube. Go to => h t t p s : / / w w w . h t t p s : / / w w w . youtube.com/watch?v=GVRKRsegiCE Also, remember the use of significant figures as well as other number adjusting

    posted on March 15, 2019
  22. Chemistry

    I would concede to your comment if the responses to the original question were correct, but they are NOT. The equilibrium constant is a numerical value of a reaction that defines its 'Extent of Reaction'; that is, how long it takes for the reactants to go

    posted on March 15, 2019
  23. Chemistry

    OK … When Σ molar volumes reactants = Σ molar volumes products => Kp = Kc Standard equation relating Kp and Kc is Kp = Kc(RT)^Δn where Δn = change in total molar volumes of reaction = Vₘ(Products) - Vₘ(Reactants). 3H₂(g) + N₂(g) => 2NH₃(g);

    posted on March 14, 2019
  24. Chemistry

    Challenge Question => When does Kp = Kc?

    posted on March 14, 2019
  25. Chemistry

    The number of each element on the reactant side must equal the number of elements on the product side. Such gives rise the the law of mass balance which states .... Sum of Mass of Reactants = Sum of Mass of Products Now, for Oxidation-Reduction Reactions

    posted on March 13, 2019
  26. Chemistry

    Just a different perspective. :-) ...............2CO(g) + O2(g) ==> 2CO2(g) I...............1490.........745.............0 C...............-2x...........-x..............2x E.................0.............0..............2x Complete Rxn => 745 + (-x) = 0 =>

    posted on March 13, 2019
  27. Chemistry

    Given 89.3-ml gas mix of O₂(g) and H₂O(g) at 21.3ᵒC (=294.3K) and 756 mmHg ( = 0.9947 atm) TTL pressure, determine the following given also P(H₂O) = 19 mmHg at 756 mmHg/21.3ᵒC). P(O₂) in mmHg, b. %V(O₂) contribution in mix & c. grams O₂(g)

    posted on March 13, 2019
  28. chemistry

    Moles CO2 = 8.8-g CO2/44g/mole CO2 = 0.20 mole CO2 x 6.02E23 molecules CO2/mole = 1.204E23 molecules CO2 in the 8.8-gram sample. #atoms Oxy per CO2 molecule = 2 Therefore, #oxy atoms in 8.8-g CO2 = 1.204E23 molecules CO2 x 2 oxy molecules/CO2 molecule =

    posted on March 13, 2019
  29. chemistry

    Not to be difficult, but the 0.3% indicates a much lower solubility for the Zn(CN)2 than the common ion calculation would support in the presence of 0.3575M Zn. The 1.448 x 10^-8M solubility is accurate with respect to the data given. A lower solubility

    posted on March 12, 2019
  30. Chemistry

    A => B ΔG = -12.9-Kj – (-11.3-Kj) = -1.6 Kj ΔG = -RT∙lnKeq => lnKeq = -(ΔG/RT) = -(-1.6Kj∙molˉ¹/0.008314Kj∙molˉ¹∙Kˉ¹∙293K) = 0.6568 Keq = exp(0.6568) = 1.928 Qeq = [2.41M]/[1.35M] = 0.5602 Keq = 1.928 > Qeq = 0.5602 => Reaction shifts

    posted on March 12, 2019
  31. Chemistry

    23.5-ml(0.12 HCl) + 50-ml(0.15M BH:OH) => 0.0235(0.12) mole HCl + 0.050(0.15) mole BH:OH => 0.00282 mole HCl + 0.0075 mole BH:OH => (0.0075 – 0.00282) mole BH:OH + 0.00282 mole BH⁺:Clˉ => 0.00468 mole BH:OH + 0.00282 mole BH⁺:Clˉ => (0.00468 mole

    posted on March 12, 2019
  32. chemistry

    DrBob, I have a different approach... Mixing 0.36M (Zn⁺²) + 0.005M (CNˉ) => 0.005M Zn(CN)₂ formed + (0.3600M - 0.005M) = 0.3575M Zn⁺² in excess => (b/c CNˉ is the limiting reactant.) ….. Zn⁺²……….+ ……… 2CNˉ…….=>..Zn(CN)₂ +

    posted on March 12, 2019
  33. chem

    I’m guessing you need the Rxn Enthalpy. It’s the only given data item missing for the 1st Law equation… ΔU = q + w => q = ΔU – w = (-1370 J) – (- 736 J) = (-1370 + 736) J = - 634 Joules exothermic heat of rxn. ΔU = -1370 Joules (given) w =

    posted on March 10, 2019
  34. Chemistry

    38.5-ml(8.9E-3M HBr) + 27.0-ml(4.6E-3M LiOH) => ?pH of final mix => 0.0385(8.9E-3)mole HBr + 0.027(4.6E-3)mole LiOH => (3.43E-4) mole HBr + (1.24E-4) mole LiOH => (2.188E-4) mole HBr (in excess)* + (1.24E-4) mole LiBr + (1.24E-4) mole H₂O *LiOH is

    posted on March 10, 2019
  35. Chemistry

    I got pH = 2.48 ... Please show some effort in solving problem.

    posted on March 10, 2019
  36. Chemistry

    I got => ∆T(f) = -6.00ᵒC, but if you want support on learning the subject you need to show some effort on the problem.

    posted on March 9, 2019
  37. Chemistry

    I got 10-Liters.

    posted on March 9, 2019
  38. Chemistry

    Formatting is tough here... The quagmire of symbols below the equation is the ICE table for the problem analysis.

    posted on March 8, 2019
  39. Chemistry

    CO(g) + H₂O(g) < = > CO₂(g) + H₂(g) P(i) 1380-torr 1730-torr 0 0 ΔP -ΔP - ΔP +ΔP +ΔP P(eq) 1380-ΔP 1730-ΔP ΔPΔP Kp = P(CO₂)∙P(H₂)/P(CO)∙P(H₂O) = (ΔP)²/(1380-ΔP)(1730-ΔP) = 0.0611 Solve for ΔP using the Quadratic Formula ... I

    posted on March 8, 2019
  40. Chemistry

    If you haven't seen this acronym, it may be helpful in identifying oxidation and reduction half-reactions... OIL RIG … O => Oxidation I => Is L => Loss (of electrons) R => Reduction I => Is G => Gain (of electrons) Alᵒ(s) + 3AgNO₃(aq) =>

    posted on March 8, 2019
  41. Chemistry

    Ya got plenty of wins! Not to worry. :-)

    posted on March 8, 2019
  42. Chemistry

    DrBob, The chemistry is correct but, where did the chromate come from? I think you meant potassium oxalate monohydrate. :-)

    posted on March 7, 2019
  43. Chemistry

    Remember, these type kinetics problems are always based upon ‘pairs’ of substances. A quick setup is to equate the rates of the two substances of interest and switch the coefficients. One of the rates will be given. Solve for the unknown in terms of

    posted on March 7, 2019
  44. Chemistry

    Convert your given mass values to moles then divide by the respective coefficients of the balanced equation. The smallest value is the limiting reactant.

    posted on March 4, 2019
  45. Chemistry

    [H⁺] = [C₃H₅O₂ˉ] = √Ka[HC₃H₅O₂]) = √(1.3 x 10ˉ⁵)(0.102) M = 0.00115M [HC₃H₅O₂] = 0.102M %Dissociation = ([H⁺] or [C₃H₅O₂ˉ])]/[HC₃H₅O₂])100% = [(0.00115M)/(0.102M)]100% = 1.13% dissociated

    posted on March 3, 2019
  46. chemistry

    Are you asking about ions formed from a given molecule? Other than that, there is similarity only in molecules and ions with similar molecular geometries based on the VSEPR or Molecular Orbital theories of molecular geometry.

    posted on March 2, 2019
  47. Chemistry

    You will need molecular weight value of the compound under-going combustion to complete the problem objective. With the data given, only the empirical formula can be found.

    posted on February 28, 2019
  48. chemistry

    Given compound with C, H, O => 4.4-g CO₂ + 2.7-g H₂O Elemental %Composition of … %C in CO₂ = (12/44)100% = 27.3% => Wt C from CO₂ = 27.3% of 4.4-g CO₂ = 1.206-g Carbon %O in CO₂ = (32/44)100% = 72.7% => Wt O from CO₂= 72.7% of 4.4-g CO₂ =

    posted on February 27, 2019
  49. General Chemistry

    A simple way to determine the limiting reactant is to convert reactant quantities into moles and then divide by the respective coefficient of each reactant of the balanced equation. The smallest value is the limiting reactant. Solve for yields based on

    posted on February 24, 2019
  50. Chemistry

    C + O₂ => CO₂ (5.25-gC/12g/mol) = 0.4375 mole C => 0.4375 mole CO₂ x 44 g/mole = 19.25 g (Theoretical Yield) %Yield = [Actual Yield / Theoretical Yield] x 100% = (10.2-g/19.25-g) x 100% ≈ 53% Yield

    posted on February 24, 2019
  51. Chemistry

    Given Conc CHCl₃ = 0.80ppb => [0.80 g CHCl₃ / 1 x 10⁹ g Water]∙100% = 8 x 10¯⁸ % w/w … Assuming 350 ml water = 350 grams... => 8 x 10¯⁸ % of 350-g = 2.8 x 10¯⁷ g CHCl₃ * =>2.8 x 10¯⁷ g CHCl₃ = 2.8 x 10¯⁷ g CHCl₃/119 g/mole =

    posted on February 24, 2019
  52. Chemistry

    At STP => moles O2 = (50L)/(22.4L/mole) = 2.232 moles O2 2.232 moles O2 = (2.232 mole O2)(32 g/mole) = 71.429 grams O2

    posted on February 23, 2019
  53. Chemistry

    Just in case you need the ion concentrations after mixing, here's my solution... Na₂SO₄ + Ca(NO₃)₂ => 2NaNO₃ + CaSO₄(s) 1:1 Rxn Ratio between Na₂SO₄ (limiting reactant) in an excess of Ca(NO₃)₂ 100ml(0.50M Na₂SO₄) + 100ml(1.65M

    posted on February 23, 2019