Doc48

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 Ca(NO₃)₂) …  0.10(0.50) mo...

6 years ago

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

6 years ago

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 = 2.353 x 10¯⁹ mole CHCl₃ => 2.3...

6 years ago

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

6 years ago

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 mo...

6 years ago

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 = 3.199-g Oxygen %...

6 years ago

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.

6 years ago

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.

6 years ago

[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

6 years ago

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

6 years ago

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 the g...

6 years ago

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

6 years ago

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

6 years ago

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) => Al(NO₃)₃(aq) + 3A...

6 years ago

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 ΔP<Equal>ΔP 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 got ΔP = 305-torr P(CO...

6 years ago

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

6 years ago

I got 10-Liters.

6 years ago

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

6 years ago

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

6 years ago

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 Limitin...

6 years ago

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 = -736 Joules (ex...

6 years ago

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 + *Unreacted excess Zn⁺² ½(0.005M)…...

6 years ago

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 BH:OH/0....

6 years ago

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 right. …………….A.........

6 years ago

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 f...

6 years ago

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 = 2...

6 years ago

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) in mix. P(O₂):...

6 years ago

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 =>...

6 years ago

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...

6 years ago

Challenge Question => When does Kp = Kc?

6 years ago

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); Δn = 2Vₘ - 4Vₘ...

6 years ago

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...

6 years ago

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 conce...

6 years ago

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...

6 years ago

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. :-)

6 years ago

Right on! Dr Bob:-)

6 years ago

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 acid is a weak acid (Kₐ =...

6 years ago

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 communit...

6 years ago

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

6 years ago

The last entry above is fm Doc48

6 years ago

Bromothymol Blue

6 years ago

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’

6 years ago

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)...

6 years ago

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 H₂Tartarate (=0.2/150 m...

6 years ago

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)

6 years ago

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 press...

6 years ago

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.

6 years ago

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

6 years ago

Great post DrBob!

6 years ago

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 CaCO₃

6 years ago

53.3g

6 years ago

6.25 mol

6 years ago

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

6 years ago

This is a 2-step process... 1. Carbon dioxide and water react to make carbonic acid. CO2(aq)+H2O(l)⟶H2CO3(aq) Carbonic acid reacts with hydroxide ions (acid/base metathesis rxn) to make sodium bicarbonate and water. 2. H2CO3(aq)+NaOH(aq)⟶NaHCO3(aq)+H2O(l)...

6 years ago

Given V(NO₃)₂(aq) + 6-amps for 800-min(=48,000-sec) a. Amps = Charge/time => amps = Coulombs/seconds => Coulombs = amps x seconds => Coulombs = (6-amps)(800-min)(60-sec∙minˉ¹) = 288,000-Coulombs ……………………………………………………………………………………… b. The Vanadium electrode...

6 years ago

Use Graham's Law of Effusion => SqrRt(M-A)/t-A = SqrRt(M-B)/t-B SqrRt(O2)/t(O2) = SqrRt(M-B)/t(B) => M-B = 63.8 amu

6 years ago

From formula => 1 mole Mg + 2 moles Cl + 8 moles O If you need the numerical number of atoms of each, remember Avogadro's Number => 1 mole contains 6.02E+23 then n-moles would be => n(Avogadro's No)

6 years ago

Your initial listing is correct... The rate depends upon the least stable carbon attached to the halogen for Sn2 reaction mechanisms. MtBr with methyl carbon is the fastest substitution EtBr and neopentyl-Br both have primary carbons attached to the Bromi...

6 years ago

If the 0.18 mole is NaClO2, I get 0.25M HClO2 at equilibrium. Note => In the process of calculating the HClO2 concentration using the ICE table analysis, Ka = 1.1E-2 which indicates by the simplification rule, one can not drop the 'x' in the equilibrium r...

6 years ago

Are you asking about 'rate determining step' in a reaction mechanism... e.g., Sn1 mechanism of t-BuBr?

6 years ago

Go to => h t t p://chemunlimited.com/Lecture%20Videos%20-%20Chap%2013%20-%20Chem%20Kinetics.html => Intro to Equilibrium (Video). remove spaces in 'h t t p'

6 years ago

Given VP(H₂O) = 42.2-mmHg @35ᵒC & Needed VP(Soln) = 40.8-mmHg VP(Solution) = VP(solvent) – ΔVP(solvent) ΔVP(solvent) = X(solute)∙VP(solvent) (42.2-mmHg – 40.8-mmHg) = [(m/180.2)/(m/180.2) + (575/18)]∙42.2-mmHg ; m = mass glucose (g) Solving for mass of gl...

6 years ago

One must 'assume' a pH value with pKa from Ka = 1.1E-2. I used pH = 1.8 (pH = 1.5 - 2.0 is typical of chloric acid solutions in the 0.1 - 0.2M range) & pKa = -log(Ka) = -log(1.1E-2) = 1.96 and got [HOCl2] = 0.25M. Same as the ICE table analysis. Great sug...

6 years ago

Yes, use mole ratios ... You also need to specify atmospheric conditions; i.e., temperature - pressure conditions for calculation of volume values. Then you can use PV = nRT to calculate volumes. One could simply assume STP, but that conglomeration you've...

6 years ago

DrBob, Correct me please, but wouldn't 5.5% w/v => 5.5-g EtOH/100-ml soln (assuming water solvent)? => (5.5-g/46.07-g/mol) / (0.100-L) = 1.2M in EtOH (2 sig. figs.)

6 years ago

At end of dive. All kinetic energy has been expended... That is, the max is while standing on the spring board and the minimum at the bottom of the dive. The extra energy content at the top of the board is converted into kinetic energy as the diver falls...

6 years ago

You are most welcome. :-)

6 years ago

I got 27.5% NaNO2 (w/w) ... post work if you get stuck. NOTE => You do need 3I^- on reactant side of the initial equation and on the product side of the thiosulfate rxn in post.

6 years ago

Your math and conclusions are correct, but suggest keeping in mind the numbers actually refer to 'molar volumes' of gas. Such could pose some confusion when working with gas volumes in practical lab settings. That is, one should specify temperature and pr...

6 years ago

The only question I would have is => why did you choose 2-sig.figs. for expressing the final answer... Typically, the final sig.figs. are used for 'practical' laboratory considerations and determination is based upon the instrument/measuring device in the...

6 years ago

In support of DrBob's suggestion, see my note on sig.figs. in previous posted problem. It's been my experience that significant figures are the one element of laboratory reports that students seem to misunderstand & misuse more than anything else. There i...

6 years ago

In support of DrBob's suggestion, see my note on sig.figs. in previous posted problem. It's been my experience that significant figures are the one element of laboratory reports that students seem to misunderstand & misuse more than anything else. There i...

6 years ago

Oops, didn't mean to post twice.

6 years ago

Here’s a repost … Determine the enthalpy of reaction for 2N₂(g) + 5O₂(g) => 2N₂O₅(g) from the following reactions using Hess’s Law. Rxn 1: H₂(g) + 2O₂(g) => H₂O(l); ΔH₁ = 2(-290)-Kj = -580-Kj Rxn 2: N₂O₅(g) + H₂O(l) => 2NHO₃(l); ΔH₂ = -77.8-Kj Rxn 3: N₂(g...

6 years ago

This is a Hess's Law Equation problem... Add the enthalpy's of formation for the products minus the enthalpy's of formation of the products.

6 years ago

Moles = Molarity x Volume in Liters => Volume (L) = moles KBR / Molarity of Solution Molarity (given) = 6-Molar in KBr = 6-moles KBr/L soln Moles KBr in 4.0-g = 4.0-g/119-g/mole = 0.034-mole KBr Therefore, Vol(L) of 6M solution containing 4-g KBr = 0.034-...

6 years ago

Math is correct, just thought I saw 6M, not 0.680M and yes the last zero should be taken as a sig fig. Apologies.

6 years ago

Here's a reference that's useful for tris systems...(remove the space between the h t, copy and paste => search) h ttp://www.sigmaaldrich.com/Graphics/Supelco/objects/4800/4709.pdf

6 years ago

Supply the missing quantum number or sub level name. n. l. ml. name (a). ? ? 0. 4p (4, 1, 0) (b). 2. 1. 0. ? () (c). 3. 2. -2. ? (d). ? ? ? 2s2s (e). 3. ? -1. 3p? n => 1, 2, 3, 4, … l => s, p, d, f, g, h … => 0, 1, 2, 3, 4, 5, 6, … m =>.......... - 3 -2 -...

6 years ago

Orders of Rxn are 'rate trends' ... That is, look at the change in rate from initial concentration to 2x, 3x, etc. the initial concentration for each reactant. The general rule is hold all concentrations constant and vary one component at a time and note...

6 years ago

This one is a gnarly scuzzard!!! Here’s my best guess… Given a solution mix containing 1M Ba(OH)₂ + excess Zn(OH)₂ … Determine pH of solution at equilibrium. Ba(OH)₂ < == > Ba⁺² + 2OHˉ ; Ksp = 3 x 10ˉ¹⁵ Zn⁺² + 2OHˉ < == > Zn(OH)₂ ; Kf = 2 x 10⁺¹⁵ ……………………...

6 years ago

Yes, you're right. it give the OH complex. Duh! Will try again. Thanks.

6 years ago

OK, here’s a rework… Zn⁺² ions in water react to form Zn(OH)₂, a gelatinous white solid. Further addition of a strong base (in this problem, Ba(OH)₂) reacts with the Zn(OH)₂ to form Zn(OH)₄²ˉcomplex that is soluble in the aqueous base solution. For additi...

6 years ago

correction ... [OH] = 4x, not 2x ... [OHˉ] = 4x = 4(3.3 x 10ˉ⁴) = 1.3 x 10ˉ³ pOH = - log[OHˉ] = log(1.3 x 10ˉ³) = -(-2.9) = 2.9 pH = 14 – pOH = 14 – 2.9 = 11.1

6 years ago

For this problem, thermal decomposition of the ammonium chloride would require an input of energy (endothermic) to dissociate the physical bond between HCl and Ammonia as an increase in disorder occurs with yield of more particles of product than the numb...

6 years ago

Ugh, in a series (period) the elements do have the same number of energy levels ... it is in groups (families) of elements that chemical properties are consistent. I would have selected 'A'. :-)

6 years ago

You question is very ambiguous. AgCl is the least soluble, Na2CO3 is the most soluble, CaSO4 & PbCl2 solubilities are in the same order of magnitude.

6 years ago

From PV = nRT = (mass/f.wt)RT => (mass/Volume) = P(f.wt.)/RT = Density P = 800 Torr = 1.053 Atm (must convert Torr to Atm b/c R-value units = L∙Atm/mole∙K) f.wt = S + 6F = [32 + 6(19)] g/mol = 146 g/mol R = 0.08206 L∙Atm/mole∙K T = 35ᵒC + 273 = 308 K D =...

6 years ago

Good idea, DrBob

6 years ago

The problem with this problem is the solubility of Al(OH)₃. With a Ksp = 4 x 10ˉ¹⁵, the solubility of Al(OH)₃ will be ~1.1 x 10ˉ⁴M in Al(OH)₃ dissolved (the rest will ppt) and the [OHˉ] = 3(1.1 x 10ˉ⁴)M = 3.3 x 10ˉ⁴ mol/L in OHˉ b/c Al(OH)₃ => Al⁺³ + 3OHˉ...

6 years ago

What makes you think it's wrong?

6 years ago

Dr Bob, After addition of 10 ml of the HCl, this looks like a 50/50 buffered system in which [OH] = Kb = 4.4 x 10^-4, giving pOH = 3.36 and pH = 10.64. Would you be so kind as to check this. Thanks, Doc 48

6 years ago

That's what I'm showing in terms of pOH = pKb too. At half equivalence point the concentration of salt = concentration of amine and cancels from the Kb expression leaving Kb = [OH] => pKb = pOH.

6 years ago

Acetic Acid(HOAc) + Sodium Acetate(NaOAc) => common ion solution that inhibits changes in pH with addition of quantities of H^+ or OH^-. If H^+ is added to the buffer solution, the excess reacts with the OAc^- ion to form HOAc and consumes the excess H^+....

6 years ago

Fm equation 2 moles H2S = 2moles(34 g/mole) = 68 g H2S => -1037 Kj => 1 g H2s => (-1037/68) Kj/g = -15.25 Kj/g

6 years ago

Magnesium is a more active metal than zinc. That is, Mg will undergo oxidation (loss of electrons) more readily than Zn which in ZnO would gain the electrons donated by Mg. The MgO bond is more stable than the ZnO bond which resists replacement by less ac...

6 years ago

Hmmm??? I thought elements were defined by their atomic numbers ( => # protons) regardless of the isotope. That is, all sulfur isotopes have 16 protons and all oxygen isotopes have 8 protons. The difference is in the number of neutrons which is not the ob...

6 years ago

All data dimensions must match the units of the Universal Gas Constant (R) being used... For your problem R = 0.08206 L-Atm/mole-K.

6 years ago

You need to check SP6 to see if it was copied correctly... possibly SF6?

6 years ago

The empirical formula and structural formula are not the same thing... The empirical formula just shows the lowest whole number ratio of elements in a compound. The structural formula is a 2-dimensional or 3-dimensional graphic formula showing a plausible...

6 years ago