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Ms Lena, the 92% dissociation for H₂S (hydrosulfuric acid)* is quiet high for an acid with a Kₐ₁ = 8.9E-8*. (*Ebbing's Gen Chem; 11th Edn.) I calculate the %ionization for a 0.76M H₂S solution to be ~0.034%. Never the less, a 0.76M H₂S solution
Dr Bob, I stand corrected… Here’s my rational… After working through the details one should get 13L O₂(g) at specified conditions. Applying standard volume – in my mind – requires converting to STP conditions then converting back to non-STP
ln(VP₂/VP₁) = ΔHᵥ/R(T₂-T₁/T₁∙T₂) substitute given data (units compatible with R = 0.008314Kj/mol-K), VP₂ = 606Torr, VP₁ = 760Torr @ T₁ = 56.5C = 329.5K, ΔHᵥ = 32.0Kj/mole and solve for T₂.
Oops! miss read question... '3'Vm of compound reacts... Dr Bob's solution is correct.
VP total (mix) = 40mmHg(given) VP(gas) = mole fraction of gas x total pressure of mix mole fraction of X(B) = 0.5/(0.2 + 0.5) = 0.7142 VP(B) = X(B) x TTL Pressure = 0.7142(40mmHg) = 28.568mmHg ~ 30mmHg (1 sig.fig)
Addendum note FYI => The ΔH value given in your problem is the ‘Standard Heat of Formation’ for POCl₃(s) from its basic elements in standard state. The Standard Heat of Formation is the heat gained or lost on formation of a substance from its basic
Converting given data to moles typically uses ... grams to moles => moles = mass(g) given/formula wt(g/mole) particles to moles => moles = particle numbers given/Avogadro's Number volume at STP (Liters) to moles => moles = Volume(L) given/22.4l/mole
Converting given data to moles typically uses ... grams to moles => moles = mass(g) given/formula wt(g/mole) particles to moles => moles = particle numbers given/Avogadro's Number volume at STP (Liters) to moles => moles = Volume(L) given/22.4l/mole
or, OIL RIG Oxidation Is Loss Reduction Is Gain
Assuming all of the carbon, hydrogen, oxygen and sulfur come from the reactants given and n-volumes means 'molar' volumes, then ... CₓHᵥSₑ + 9O₂ => 3CO₂ + 6H₂O + 3SO₂ => C₃H₁₂S₃ + 9O₂ => 3CO₂ + 6H₂O + 3SO₂ => Empirical Formula
Oops! => correction on equation balance... 2C4H10 + 13O2 => 8CO2 + 10H20
Dr Bob, Check that Oxygen coefficient. => 13 Oxy => 2(13/2)L = 13L Oxy in rxn by your analysis. :-) => 2C4H10 + 13O2 => 4O2 + 5H2O Another approach is to convert the 2L of Butane to STP volume and divide by 22.4 L/mole. This gives from rxn stoichiometry =>
Correction... Total Pressure = 90 + 45 = 135mmHg VP(MtOH) in mix = X(MtOH)∙VP(MtOH) = 0.50(135mmHg) = 67.5mmHg VP(EtOH) in mix = X(ETOH)∙VP(EtOH) = 0.50(135mmHg) = 67.5 mmHg TTL Pressure = 135mmHg
VP(MtOH) = 90mmHg m(MtOH) = 64g/32g/mole = 2 mole VP(EtOH) = 45mmHg m(EtOH) = 92g/46g/mole = 2 mole X(MtOH) = X(EtOH) = 2/(2 + 2) = 0.50 VP(MtOH) in mix = X(MtOH)∙VP(MtOH) = 0.50(90mmHg) = 45mmHg VP(EtOH) in mix = X(ETOH)∙VP(EtOH) = 0.50(45mmHg) =
Notation => [Oxidation Rxn||Reduction Rxn] [Pt(s) | H₂(g) | H⁺(aq) || SO₄ˉ²(aq) | PbSO₄(s) | Pb(s)] H₂(g) + Pt(s) => 2H⁺(aq) + 2eˉ (Oxidation Half Rxn) PbSO₄(s) + 2eˉ => Pb⁰(s) + SO₄ˉ²(aq) (Reduction Half Rxn)
Litmus paper is an indicator for the presence of an acidic solution (blue litmus => red) or alkaline solution (red litmus => blue). To understand this it is important to note that for a substance to become an acid or base two conditions must apply; that
Ksp expressions do not have the solid form of the substance of interest in the equilibrium expression. The concentration of salt is infinitely larger than the concentration of ions delivered into solution and is therefore considered a constant. This
Kc = [CO][H₂]²/[CH₃OH] ... substitute given data and solve for Kc.
Litmus paper is an indicator for the presence of an acidic solution (blue litmus => red) or alkaline solution (red litmus => blue). To understand this it is important to note that for a substance to become an acid or base two conditions must apply; that
The ratio of C-12 to C-14 as a function of remaining activity => age of artifact. - t∙k = ln(C-14/C-12); t = age of artifact Note FYI: For living organic tissue, C-14 activity = C-12 activity. For dead organic tissue, C-14 activity < C-12 activity. At
Raoult’s Law => VP(Soln) = X(Solvent)∙VP(Solvent) = VP(Solvent) – X(Solute)∙VP(Solvent)
Boron has only 3 valence electrons. => B:[He]2s²2p¹ + 3F:[He]2s²3p²p²p¹ => B:[He](sp²)¹(sp²)¹(sp²)¹ + 3F:[He]2(sp³)²(sp³)²(sp³)¹ => BF₃:[He]2(σ∙σ∙σ) => Trigonal Planer Geometry with 3 covalent σ-bonds with θᵦ = 120⁰.
CuCl₂(aq) => Cu⁺²(aq) + 2Clˉ(aq) Clˉ(aq) + H₂O(l) => No Reaction (Theoretically Clˉ + H₂O => HCl + OHˉ. Since HCl is a strong acid and remains 100% Ionized this hydrolysis does not occur.) Cu⁺²(aq) + H₂O(l) => Cu(OH)₂(aq) + H⁺(aq) (Kf
Oops! ΔT(metal) = 25.4⁰C – 99.5⁰C
Use Σq = 0 = (m∙c∙ΔT)water + (m∙c∙ΔT)metal Water m = 125ml = 125g (1ml H₂O = 1gm H₂O) c = 1cal/g∙⁰C or 4.184j/g∙⁰C ΔT = 25.4⁰C – 22.0⁰C Metal m = 2.36102g c = ? ΔT = 99.5⁰C – 22.0⁰C (m∙c∙ΔT)water +
Assuming all of the 6.5 grams P₄ is consumed in an excess of O₂ to produce 11.54 grams PₓOᵥ then … %P per 100-wt = (6.5g/11.54g)100% = 56.3258% => 56.3258g/30.9738g/mol = 1.8185 mole P %O per 100-wt = 100% - 56.3258% = 43.6740% =>
Pressure (P) is directly proportional to height difference (h). Therefore … P ∝ h => P = k∙h => k = P/h k₁ = k₂ => P₁/h₁ = P₂/h₂ => h₂ = h₁(P₂/P₁) = 32.6mmHg(715.7 Torr/764.7 Torr) = 30.5mmHg height differential after pressure
Salicylic Acid + Acetic Anhydride.......=>..........Aspirin + Acetic Acid …..C₇H₆O₃…….. + …..C₄H₆O₃……… =>……...C₉H₈O₄ + HC₂H₃O₂ 2g/(138g/mol)….+.....(excess)………=> ?g(Theoretical) + ---------- =0.015mole
0.50g U-235 => (2.13E-3 mole)(6.02E23molecules/mole)(235nuclei/molecule)(7.59MeV/nucleon)* = 2.287E24Mev = (2.287E24MeV)(1.60218E-16Kj/Mev) = 3.66E8Kj per 0.50g U-235. *7.59MeV/nucleon = mass defect for U-235 mass to energy conversion equivalence. It is
A solution with a pH = 5.45 would in fact have a hydronium ion concentration of 3.55 x 10ˉ⁶M, but for a 0.050M NH₄Cl(aq) solution the [H⁺] is closer to 5.27 x 10ˉ¹⁰M, pH = 5.28 and Ka(NH₄⁺) ~ 5.56 x 10 ˉ¹⁰. NH₄Cl(aq) => NH₄⁺ +
From given reaction equation, moles acid consumed = moles H2 produced = 19g/2g/mole = 9.5 moles acid = Molarity x Volume(L) = 6M x Vol(L) => 1.583 Liters of the Acid Solution
Heating Iron(II)Sulfate Heptahydrate (Cuprian Melanterite)* to 110⁰C will only decompose the hydrate into the anhydrate crystalline solid Iron(II) Sulfate and water vapor. Decomposition of the FeSO₄(s) occurs ~680⁰C into Iron(III) Oxide and a mix of
2MnO₄ˉ + 4H₂O + 6eˉ => 2MnO₂ + 8OHˉ ….. E⁰ = + 0.83 volts ……………..……..6Iˉ => 3I₂ + 6eˉ …….…… E⁰ = + 0.54 volts ………………………………………………………………………… 2MnO₄ˉ + 6Iˉ +
Boyles Law P₁∙V₁ = P₂∙V₂ V₁ = 70ml P₁ =760mmHg V₂ = 16ml P₂ = ? T₁ = T₂ = 22⁰ = 295K = Constant Substitute given values and solve for P₂. P₂ = 3325mmHg = 4.375Atm or, 1 sig.figs. (V₁ = 70ml) => 3000mmHg = 4Atm
%Yield = [Actual Lab Yield mass/Theoretical Calculated Yield mass]100% Theoretical Yield calculations are based upon the limited reactant in the problem; that is, in the posted problem the limiting reactant is the iron (Fe) with Oxygen remaining in excess
If one titrates a buffered solution and a non-buffered solution where both have the same concentration of weak electrolyte (wk acid or wk base), both would have the same equivalence point. The difference is on addition of titrant (e.g., NaOH into HOAc/OAc
HO-CH₂ CH₂ CH₂-OH < CH₃CH₂ CH₂ CH₂-OH < CH₃CH₂-O-CH₂ CH₃ Hydrogen Bonding decreasing => Vapor Pressure increasing The -OH group promotes H-Bonding.
None will form ionic bonds as a metal + nonmetal are required to generate the ionic compound. HF, HCl & HBr are polar covalent molecules with HF having the higher polarity followed by HCl and lastly HBr. However, it's interesting to note that (if in
Is the 650 grams the solvent-water or cold pack? It's the water temperature that drops, not the mass of cold pack. So, I'm assuming the mass value is that of water into which the cold pack was added. Also, when measuring changes of temperature, one is
Published is ~200 MeV per nucleon fission energy of U-235 => 200 MeV/nucleon x (0.50/235)mole x 6.023E+23 nucleons/mole = 2.6E+23 MeV for fission of 0.50g U-235. Not sure why ΔE = mc² does not match this quantity. Both calculations are mass defect
ΔE = mc² … m = 0.50g = 5 x 10ˉ⁴ kg … c = 3 x 10⁸ m/s (speed of light in vacuum) ΔE = (5 x 10ˉ⁴ Kg)(3 x 10⁸ m/s)² = 4.5 x 10¹³ kg∙m²/s² = 4.5 x 10¹³ joules = 2.8 x 10²⁶ MeV
Standard conditions refer to the ‘Thermodynamic Properties of Substances’ table which is at 25ᵒC & 1 Atm. That is, calculation of ΔGᵒ, ΔHᵒ & ΔSᵒ are based on the values in that table. Then apply the results to the Gibbs Free Energy equation
V(RMS) in meters/sec = 158[SqrRt(T/M)]; T is in Kelvin, M = Mole weight.
A physical change does not alter the chemical nature of the substance. Freezing of water to form ice (solid state) H2O(l) => H2O(s) only changes state of existance but not its chemistry. A chemical change alters the chemical nature of the substance
Google it... 'Hydrogen Bonding' and read!!
For making a solution with predefined volume and concentration from a solid stock reagent (assuming 100% pure chemical), the formula ... grams reagent needed = Molarity needed x Volume needed (Liters) x Formula Wt of reagent So, for your problem ... given
Choice 3 is an incorrect statement... Buffer solutions 'limit' changes in pH by common ion effect when acid or base is added, they do NOT prevent changes.
For aqueous electrolysis of AgNO₃, water would dominate at both electrodes. Generally, for an aqueous electrolytic process whose chemical reactions are driven by a battery … The more negative Reduction Potential dominates at the cathode (site of
1 Atm supports a column of mercury 760 mm high, so, A column of 1.10 m = 1100 mm => P = 1100 mm/760 mm/Atm = 1.5 Atm (2 Sig. Figs)
Σq = q(Ag) + q(H₂O) = (mcΔT)Ag + (mcΔT)H₂O = 0 => m(Ag)(0.240j/g∙ᵒC)(26.6-59.8)ᵒC + (100g)(4.184(0.240j/g∙ᵒC)(26.6-24.6)ᵒC = 0 => m(Ag)(-7.968j/g) + (836.8j) = 0 => m(Ag) = (836.8/7.968)g = 105.2g
