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DrRebel
Answers (103)
For weak acids in pure water ... pH = -log[H^+] [H^+] = SqrRt[(Ka)(Molar Conc of Acid)] HClO: [H^+]=(SqrRt[(10^-4)(0.055)])M = 0.0024M pH=-log(0.0024)=-(-2.63)=2.63 HCOOH: [H^+]=(SqrRt[(10^-8)(0.110)])M =3.32E-5M pH=-log(3.32E-5)=-(-4.50)=4.50
Using your numbers, looks like Kc < Qc. Rx shifts from product side to reactant side. Shouldn't the PCl5 increase and PCl3 & Cl2 decrease?
Boron is not a choice. Beryllium perhaps (Be).
D: 2 core electrons The valence level in an electron configuration is the highest principle quantum number. Using the Afbau symbology that would be highest Principle Quantum Number for the element in question => in this case n = 2. The 1s² is the core
Also, check assay label for purity. If less than 100% pure, divide grams by fractional purity.
C6H12O6 + 6O2 => 6CO2 + 6HOH 5g C6H12O6 = (5/180)mole C6H12O6 = 0.028mole C6H12O6 Fm equation ... if ... 1 mole C6H12O6 => 6 moles CO2 then 0.028 mole C6H12O6 => 6(0.028) mole CO2 = 0.468mole CO2 => grams CO2 = (0.468mole CO2)(44g/mol) 20.6g CO2(g)
Adding 100-g HOH = 100-ml HOH into 10.6-M NaOH => 10.6 mole/Liter + 0.100L HOH => 10.6 mole NaOH/1.10 Liters Solution => 9.64 Molar in NaOH. H2SO4 + 2NaOH => Na2SO4 + 2HOH 1/2(9.64M) H2SO4 + 9.64M NaOH => 1/2(Na2SO4) + HOH => (4.82mol/L)H2SO4 +
(8 x 7.13)/(2/108) = ________ Kj/mol exothermic
Also, when writing formulas of ionic compounds, the 'polyions' such as the nitrate ion (NO3-), sulfate ion (SO4)-2 and phosphate ion (PO4-3) need to be in parenthesis so the subscripts outside the parenthesis indicate that you only have '2' ions of the
See your 8:14pm post.
OK... When writing formulas of neutral compounds the sum of the charges of the ions (monatomic or polyatomic) must equal zero. NaCl is 1:1 ratio b/c Na => +1 and Cl => -1. (+1)+(-1)=0. Now, Strontium is a Group II cation which always has a +2 charge in
This is a common ion problem. The C(initial) should have .0035M under CrO4^-2 which carries to E-line where 'x' can be dropped b/c it's too small to make a difference. The rest is perfect.
Convert gms to moles (divide given mass in grams by formula weight) and use equation ratio... f.wt.(KOH) = 56 g/mole 2gms KOH = (2/56)mole KOH = 0.036mole KOH fm equation 6molesKOH => 1moleKClO3 then ... (6/0.036) = (1/x) => x = (0.036/6)mole KClO3 = 0.006
I'm assuming the CH2 should be C2H2 (Acetylene) ... From the balanced equation, two moles of acetylene => -2511Kj = 2.511Mj Convert 100g C2H2 to moles 100g/26g/mol = 3.85 moles Set ratio ... (2/3.85 = 2.511/x => x = [(2)(2.511)]/(3.85)Mj = 1.30Mj
∆U = ∆H + P∆V If P∆V is negative => ∆U = ∆H + (- P∆V) => ∆U = ∆H - P∆V For work = - P∆V the particles of an expanding gas will lose energy => exothermic and will perform work on the surroundings. You can generalize this … expansion
4g NaOH = 0.10 eqv. Wt. If Eqv Wt Acid = Eqv Wt Base then => Neutralization Choices 1,2 & 4 are all 0.10 eqv wt. Therefore, 1,2 & 4 will neutralize 0.10 eqv wt NaOH. Eqv Wt = Normality x Volume in Liters Choice 3 is also a strong alkali base. Base + Base
λ = h/p (deBrogle Equation) h = Plancks Constant = 6.63 x 10⁻³⁴ Joule-sec λ = 0.10nm = 1 x 10⁻⁸ meter p = particle momentum = h/λ = (6.63 x 10⁻³⁴ Joule-sec / 1 x 10⁻⁸ meter) = 6.63 x 10⁻²⁶ Kg·m/sec
Thank you BP for the link. I'm passing it on to my students. Good Job!
Another way to look at this is ... changes in volume are pressure and temperature sensitive... => Increasing Pressure => Decreasing Volume and Decreasing Pressure Increasing Volume. => Increasing Temperature => Increasing Volume and Decreasing Temperature
2Al + 6HCl => 2AlCl3 + 3H2 (1.35/27)mole Al = 0.05molAl => 3/2(.05)mol H2 = 0.075mol H2 Vol H2 = nRT/P = [(0.075)(0.08206)(294)(760)]/(743) = 1.85L H2
The answer is 'C' choice. In freezing pt depression and boiling point elevation the number of ions delivered into solution must be multiplied by the molality and constant. This is referred to as the van't Hoff factor. ∆T = (Kb)(m)(i) ∆T = Boiling Pt
The Normality of H3PO4 (f.wt. = 98 g/mol) depends upon what type of reaction it is associated with. If it is in a metathesis reaction, the equivalent weight is the mass of reagent delivering 1 mole of + charge. => H3PO4 => 3H^+ + PO4^- => Eqv Wt =
Correction ... The [Cr^+3] is 0.039M & the [IO3^-] = 3x = 3(0.039M) = 0.117M. Sorry bout that.
After adding 75ml of NaIO3 into 125ml of 0.005M Cr^+3 => [Cr^+3]=[(0.125L)(0.005M)/(200ml Soln) => [(0.125L)(0.005M)/(0.200L)] = [(0.000625mole Cr^+3)/(0.200L)] = 0.00313M in Cr^+3 Ksp = 5x10^-6 = [Cr^+3][IO3^_]^3=(0.00313)(3x)^3 =>5x10^-6 =
----NH3 + HOH => NH4OH(fm NH4NO3) -----NH4OH NH4^+ + OH^- I:-->0.50M --------0.20M-----0 C:---(-x)---------- (+x)----(+x) E:--(0.5-x)--------(0.2+x)---(x) ---->(~0.5)-------->(~0.2)---(x) Kb=[NH4^+][OH^-]/[NH4OH] 1.8X10^-5= (0.2)(X)/(0.5) Solve for 'x'
1 mole => 6.02x10^23 molecules X => 7.50x10^24 molecules
Hypothesis is the core of the Scientific Method. After studying the problem in the literature, one comes up with a theoretical method in which to solve the problem. Some refer to the hypothesis as an 'Educated Guess as to how to solve the problem... I
Not! To much and in wrong units.
No one is going to answer a set of MC questions for you.
Assuming the process is at STP, 85gms N2 = 3.04 moles N2 => 2(3.04 moles NH3) => 2(3.04 moles)(22.4L/mol)NH3.
How much OH in pH=5 and How much OH in pH=6.5. Determine difference and that's how much CaO is needed.
You may also wish to see my post on the question from 'Urgent' at 7:02. Just before your post. Same kind of question. Ya'll must be in the same class.
Remember, LeChatlier's Principle says that if a stress is applied to a reaction the reaction will naturally shift away from the applied stress to remove it. This will continue until a new equilibrium is established. For your problem, in order increase the
Adding Cl2 would increase the stress on the reactant side as an excess to the equilibrium. To relieve the stress the reaction would shift to the right decreasing PCl3 which would be reacting with the excess Cl2 until a new equilibrium is established.
The only way to report the yield with the data given is in 'general' terms of number of molar Volumes (nVm). BP is correct, you will need Temp and Pressure values to calculate specific volumes. N2 + 3H2 => 2NH3 85gms N2 = 3.04mole N2 => 2(3.04mole NH3) =
Hope ya'll don't mind, but I have a little bit of a different approach to this kind of problem... Anyways, here's my take... The answer for ‘a’ is correct, but for ‘b’, I compute a different results… Here’s my take on these problems… Hope it
You are correct about the buffer salt being consumed and that there are 0.001 mole of HOAc remaining. Now put this in 12 ml solution and calculate the [HOAc] for the resulting solution. [HOAc] = (0.003mol/0.012L) = 0.083M. Given Ka = 1.8x10^-5 & [HOAc] =
20ml(0.150M HCl) + 10ml(0.0300M NaOH => 0.003mole HCl + 0.003mole NaOH => (0.003mol/0.03L)HCl + (0.003mol/0.01L)NaOH => 0.30M HCl + 0.30M NaOH => 0.30M NaCl + 0.30M HOH Since neither Na^+ or Cl^- will undergo hydrolysis (rxn with water), the outcome of the
Cf = Ci(e^-kt) Ci = 98.4 gms Cf = ? k = (0.693/3.8)days^-1 = 0.8663 da^-1 t = 19 days Cf = Ci e^-kt Cf = 94.4gms e^-[(0.8663da^-1)(19da)] Cf = (94.4)(7.11 x 10^-8)gms Cf = 6.7 x 10^-6 gm remains
did this work for you?
If a diamond is pure carbon, then 1 mole of carbon = 12 gms carbon = 12 grams diamond = 6.02 x 10^23 atoms of carbon => (6.02 x 10^23 atoms / 12 gms) = (5.02 x 10^22 atoms / gm diamond)x(0.20gm/carat) = 2.51 x 10^23 atoms of diamond-carbon/carat
Given: k₁ = 1.1 x 10¯ˢ, T = 25⁰C = 298 K k₂ = 2.4 x 10¯², T = 225⁰C = 498 K Arrhenius Equation: ln(k₂/k₁) = [(∆Eₐ/R)((T₂ - T₁)/(T₁·T₂))] Solve for ∆Eₐ ∆Eₐ = [R· ln(k₂/k₁)·( T₁·T₂)]/[T₂ - T₁] ∆Eₐ =
Zn⁰(s) + Cu⁺²(aq) => Zn⁺²(aq) + Cu⁰(aq) Oxidation => Zn(s) => Zn⁺²(aq) + 2e¯ (The loss of 2e¯ means Zn⁺²(aq) has moved into solution leaving 2e¯ behind in the Zn(s) electrode which are responsible for the anodic negative charge on the
If you can figure out the non-standard cell potentials for your problem, then substitute in the Nernst Equations at the end of this dialog... For [Zn⁰(s)│Zn⁺²(aq)ǁSn⁺²(aq)│Sn⁰(s)] Half-Cell 1 => Zn⁰(s) => Zn⁺²(aq) +2e¯; E⁰ = -
Sabrina, This is probably way more than you need, but you seem very anxious about this concept when it is quite simple in theory. Please don’t stress out over something like this. It’s really bad for you social life… Anyways, Here’s my take on the
[NaOH] = [(25g)/(40g/mole)]/(0.400Liters) = 1.56M
When did Carbon get 7 protons?
A. Mole Fraction (X₁) = [ moles of substance(n₁) / Total moles of all substances (∑n) Moles H₂ = (6 gms/ 2 gms/mol) = 3 moles Moles N₂ = (7 gms/ 14 gms/mol) = 0.50 mole ∑moles = (3 + 0.50)moles = 3.5 moles X(H₂) = [(moles H₂)/(moles H₂ +
Sorry, used 1.50 gms Na instead of 1.15 gms Na. Just change the 1.5 to 1.15 and the problem is the same. [Na⁰(s)]added = [NaOH]formed = [OH¯]ionized = 1.15g/23g-mol¯¹ = 0.05 M OH¯ pOH = -log[OH¯] = -log(0.05) = 1.3 => pH = 14 – pOH = 14 – 1.3 =
Rxn: Na⁰(s) + H₂O(l) => NaOH(aq) + ½H₂(g) Moles Na⁰(s) = (1.50 gms / 23 gms/mol) = 0.065 mol Na⁰(s) From Rxn ratios … Moles Na⁰(s) consumed = Moles NaOH formed => 0.065 mol Na⁰(s) => 0.065 mol NaOH(aq) => [NaOH] = (0.065 mole NaOH / 0.500
