DrRebel

Fe(OH)3 <=> Fe^+3 + 3OH^- Ksp = [Fe^+3][OH^-]^3 Ksp = (x)(3x)^3 = 27x^4 => x = Solubility of Fe(OH)^3 ---- = (Ksp/27)^1/4 ---- = (1x10^-39/27)^1/4 ---- = 7.8x10^11 Molar [OH^-] = 3(7.8x10^-11)M = 2.4x10^-10M => pOH = -log[OH^-] = -log(2.4x10^-10) = 9.63 p...

9 years ago

SAFETY NOTE WHEN USING/MIXING STRONG ACID SOLUTION IN WATER: Add X-ml of HCl needed into a small quantity of water ~ 40-ml or so, then dilute to needed volume. Adding water directly into 18 ml of Concentrated HCl will generate so much heat that splash-bac...

9 years ago

6Fe^+2 => 6Fe^+3 + 6e^- (Oxidation; Fe^+2 is the Reducing Agent) Cr2O7^2- + 2(3e^-) => 2Cr+3 + 7HOH (Reduction; Cr2O7^2- is the oxidizing agent) ---------------------------------- Net Redox Rxn: Cr2O72- + 6Fe^+2 + 14H^+ = 2Cr^3 + 6Fe^+3 + 7HOH

9 years ago

for more DrReb048(at)g m a i l (dot) c o m

9 years ago

for more DrReb048(at)g m a i l (dot) c o m

9 years ago

(Redn half-rxn; Cr2O7^2- is Oxidizing Agent) 16HOH + 2K^+ + 4Cr2O7^2- + 12e^- => 4Cr2O3 + 20OH^- (Cr^+6 => Cr^+3 + 3e^-) (Oxidation half-rxn; S^o is the Reducing Agent) 12OH^- + 3S^o => 2K^ + 3SO2 + 6HOH + 12e^- (S^o => S^+4 + 4e^-) Net Oxdn-Redn Rxn (K^+...

9 years ago

The electrodes in both Electrolytic and Galvanic Cells are defined by the chemistry that occurs at a given electrode. The anode is always the site of oxidation and the cathode is the site of reduction. In electrolytic cells, the electrodes are connected t...

9 years ago

There are also some great links online that illustrate the processes in both types of cells. DrReb048(at)g.m.a.i.l(dot)c.o.m

9 years ago

Convert grams to moles, then divide moles by respective coefficients in balanced equation. The smaller value is the limiting reagent and the other(s) are in excess.The amounts used are proportional to the limiting reagent. 2H2 + O2 => 2H2O moles H2 = 5g/...

9 years ago

Also, sum of valence values must = 0... Ca^+2 + 2(OH^-) => 1(Calcium Ion Valence) + 2(Hydroxide Valence, or net charge) = 1(+2) + 2(-1) = 0 and combination of elements constitute a correct formula ratio.

9 years ago

The common manufactured stock of HCl from commercial producers is 35% and has a density of 1.175 g/ml. Practical applications typically extract a specific volume of the concentrated HCl and dilute to a desired volume. A more practical problem would read.....

9 years ago

Factors affecting rxn rates are: C.A.N.T.C C => Concentration Effects A => Area (Surface Area) N => Nature (Structural Chem of Reagents T => Temperature C => Catalysts The relative effect of one factor over the other varies with the desired outcome of the...

9 years ago

A) H2O would most likely donate the electron pair from oxygen into Carbon-5 as it is carrying a +1 formal charge (not shown in question). CH3CH2CH=CH-CH2(+) + H:O:H (which carries 2 non-bonded electron pairs) The Pi-electrons of the double bond between C-...

9 years ago

Right on the money, BP... If I may add, how about an acidic 50/50 buffer. HOAc/NaOAc (Acetic Acid/Sodium Acetate), Ka=1.8x10^-5... For any 50/50 acidic buffer given the Ka of the weak acid, the buffer pH = pKa = -log(Ka) = -log(1.8x10^-5) = -(-4.74) = 4.7...

9 years ago

It might be instructive to note that ALL radio decay is 1st order decay. Such is the value of the 1st order decay equation C(remaining) = C(initial)e^(-kt) where k = rate constant = 0.693/(Half-Life). In this case Half-Life of C-14 = 5730 Years => k = (0....

9 years ago

Solution Prep'n by Dilution of Lab Stock ... (Molarity x Volume)concentrated soln = (Molarity x Volume)diluted soln M(conc) = 1.25M V(conc) = ? (= volume needed to dilute to 50 Liters.) M(dild) = 0.05M V(dild) = 50 L [(1.25M)(X) = (0.05M)(50L) (x) = [(0.0...

9 years ago

Mg(OH)2 + 2HNO3 => Mg(NO3)2 + 2HOH moles = Molarity x Volume(L) moles HNO3 = (2.5M)(0.385L) =(2.5mol/L)(0.385L)= 0.963mole HNO3 From balanced equation, half as many moles of Mg(OH)2 are needed as the amount of HNO3 given. That is, moles of Mg(OH)2 needed...

9 years ago

1.35gms. From the Ideal Gas Law PV=nRT solve for moles CH4 = n = PV/RT = (1-Atm)(2.00-L)/(0.08206-L-Atm/mol-K)(273+15)K = 0.085-mole CH4... Convert to grams by multiplying by formula wt of CH4 = (0.085-mol)(16-gms/mol) = 1.35 gms.

9 years ago

KOH + HCl => KCl + H23^+

9 years ago

molality = moles solute/Kg Solvent It is a concentration term used mainly for systems that go through a change in temperature and need a constant value for concentration. Concentrations in terms of Molarity would change with changes in temperature due to...

9 years ago

VP(Soln) = VP(Solvent) - VP(Lowering Factor) Using the following: => VP(Lowering Factor) = VP(LF) => VP(Solvent) = VP(Solv) => Mole Fraction Solute = X(Solute) = X(Solu) => van 't Hoff factor = (VHF) VP(Soln) = VP(Solv) - VP(LF) --VP(LF-ionic Soln) = VP(S...

9 years ago

OK, that makes sense... The ionic equation.

9 years ago

Where in the world does one call sulfuric acid 'Tetraoxosulphate6 Acid'? Anyways ... Molecular Equation ... H2SO4 + 2KOH => K2SO4 + 2HOH Ionic Equation ... 2(H^+) + (SO4^2-) + 2(K^+) + 2(OH-) => 2(K^+) + (SO4^2-) + 2(HOH) (K^+) and (SO2^2-) are spectator...

9 years ago

What indicator are you using?

9 years ago

This equation needs reviewing. Both PbCl2 and PbBr2 are basically insoluble with very little probability of reacting as listed. The concentration of Pb^+2 is in the order of 10^-5M which translates to consuming only 2 x 10^-5M in Na-Halide. Kinetic feasib...

9 years ago

It's a bit tough showing the oxidation number method in this format. Here's the half reaction format... 2Al^o => 2Al^+3 + 2(3e^-) 3Pb^+2 + 3(2e^-) => 3Pb^o --------------------------- 2Al^o + 3Pb^+2 => 2Al^+3 + 3Pb^o

9 years ago

Spontaneous Process is reaction without having to input energy. Spontaneous reactions will proceed to products automatically, typically releasing energy. Rusting of Iron is an example of chemical spontaneity. 2Fe^o + 3O2 => Fe2O3(s) => Rust. Also, your ca...

9 years ago

Your question needs more information about the chemical process under consideration.

9 years ago

1st, the Standard Enthalpy values listed are for solid salts and not aqueous salts. These values can not be used in aqueous solution phase reactions. 2nd, the calculation of Enthalpy of reaction of a salt(aq) + salt(aq) => salt(aq) + ppt(s) reaction shoul...

9 years ago

Here's a head-to-tail condensation I remember from my Grad School days; but after this, you'll have to entertain contributions from others... Bis-Phthaloyl Chloride + Ethylene Glycol (typically in an aprotic solvent) => Polyphthaloylethylene Glycol Ester...

9 years ago

Here's a head-to-tail condensation I remember from my Grad School days; but after this, you'll have to entertain contributions from others... Bis-Phthaloyl Chloride + Ethylene Glycol (typically in an aprotic solvent) => Polyphthaloylethylene Glycol Ester...

9 years ago

That is correct (100%)... The anion from the salt bridge migrates to the anodic cell to off set the build up of cations from oxidation and the cation from the salt bridge migrates to the cathodic cell to offset the loss of cations being reduced to basic s...

9 years ago

The valence shell of Carbon (forming CH4), Nitrogen (forming :NH3) and Oxygen (forming HOH with 2 non-bonded pair) undergoes hybridization into 4sp^3 Hybrid Orbitals which look like skewed figure 8s... This is like taking the valence shell orbitals and pu...

9 years ago

HOH => 2 + 8 + 2 = 12 e^-/molecule HOH => (12 e^-/molecule)(6.02x1-^23)molecules/mol))(1mol/18g)(1g/ml) => ([(12)(6.02x10^23)(1)(1)]/[18]) => 4x10^23 e^-/18-ml HOH

9 years ago

Given 0.311g sample = 1.50 mmole sample = 1.5x10^-4 mole sample Sample mole weight = (0.311)g/(1.5x10^-4)mol = 207.33 g/mol fwt(Xe) + n·fwt(F) = 207.33 g/mol (131.3)g/mol + n·(19.0)g/mol = 207.33g/mol (n·19)g/mol = (207.33 - 131.3)g/mol n = [(207.33 - 131...

9 years ago

? gms CO2 from decomposition of 200 gms of CaCO3. CaCO3 => CaO + CO2 moles CaCO3 = (200g/100g/mol) = 2 mol 2mol(CO2) => 2mol(CaO) + 2mol(CO2) 2mol(CO2) = 2mol(CO2)(44g/mol) = 88gms CO2.

9 years ago

A) CH3(CH2)4COH + NHR2/Hydride Reducing Agent => CH3(CH2)4CH2NR2 B) CH3CH2CH2COOH + KOH => CH3CH2CH2COO^-K^+ + HOH

9 years ago

You can also use for weak acids in pure water the 'Sqr Root' formula... [H^+] = Sqr=Root(Ka·[weak acid]) [H^+] = Sqr=Root[(4.5x10^-4)(0.90)]M = 0.02M for weak bases... [OH^-] Sqr=Root(Ka·[weak base])

9 years ago

You are most welcome... DrReb(at)g.m.a.i.l.c.o.m

9 years ago

The extra info could be a common ion relative to the Chromium(III)Iodate, but plugging 0.005M Cr^+3 into the equilibrium does not give a solubility that matches any of the answer choices. The only answer consistent with some of the data is solubility of C...

9 years ago

0.186M(KI) => 0.186(K^+) + 0.186(I^-) ------- AgI <=> Ag^+ + I^- C(eq) ----------x-----0.186M Ksp = [Ag^+][I^-] = (x)(0.186M) Ksp(AgI) = 8.3x10^-17 (fm table of Ksp values) = (x)(0.186M) x = Solubility in presence of 0.186M(KI) = (8.3x10^-17/0.186)M = 4.5...

9 years ago

A) pH of Buffer Mix: (HO)3-CNH4OH <=> (HO)3-CNH4Cl + OH [(HO)3-CNH4OH] = [(10g/97g/mol)/(0.250L)=0.412M [(HO)3-CNH4OHCl] = [(10g/115g/mol)/(0.250L)=0.348M Kb(fm table of wk base ionization constants) = 1.202x10^-6 => Ka = Kw/Kb = (1.0x10^-14/1.202x10^-6)...

9 years ago

From table of thermodynamics constants, the equilibrium constant can be calculated using Thermodynamic constants for Free Energy of Formation(∆G⁰fmn) to calculate Free energy of reaction(∆G⁰Rxn). Substitute (∆G⁰fmn) values into Hess's Law Equation for Fre...

9 years ago

Awh Shucks! It wern't nut'n!

9 years ago

More help => [email protected](dot)c.o.m.

9 years ago

For salts in pure water & with low solubilities, the molar amount of compound delivered into solution can be calculated from simple formulas based on ionization ratios ... Salt--(Ionz'n Ratio)--Solubility(S) AB => 1:1 => S = (Ksp)ª; a = ½ AB₂ & A₂B => 1:2...

9 years ago

2 x 10⁻⁵mole HCl in 500-ml Solution: =>[HCl] = [H₃O⁺] = [H⁺] = (moles H⁺ / volume in Liters) = (2 x 10⁻⁵ mole H⁺/ 0.500 Liters) = 4.0 x 10⁻⁵ Molar in H⁺ pH = -log[H₃O⁺] = -log[H⁺] = -log(4 x 10⁻⁵) = 4.4 Diluting to 1.5 Liters: => [HCl] = [H₃O⁺] = [H⁺] = (...

9 years ago

If one assumes that the 22.41mL of 0.163M NaOH(aq) was used to titrate through both equivalence points, then 2 moles of NaOH(aq) are used for each mole of H₂C₂O₄(aq) neutralized. That is… H₂C₂O₄(aq) + 2NaOH(aq) => Na₂C₂O₄(aq) + H₂O(l) 2 moles NaOH > 1 mol...

9 years ago

You need to make a plot of 'undecayed' M&M's vs Throw number. Undecayed M&M's = (UD) => UD(1) = (Total M&M's Before 1st throw) - (M's-up after 1st throw) UD(2) = [UD(1) - (M's up after 2nd throw)] UD(3) = [UD(2) - (M's up after 3rd throw)] UD(4) = [UD(3)...

9 years ago

Let n = moles and V = Volume n ∝ V => n(1)/n(2) = V(1)/V(2) n(1) = (0.833 gms / 4 gms/mol) = 0.2083 mol n(2) = [(0.833 – 0.203) gms / 4 gms/mol) = 0.1575 mol V(1) = 18.3 Liters V(2) = X (0.2083 mol / 0.1575 mol) = (18.3 Liters / X) X = [(0.1575 mol)(18.3...

9 years ago

Q = m·∆H(fusion) = 75.0 gms (80 calories/grm) = 6000 calories = (6000 calories)(4.184 joules/calorie) = 25,104 joules

9 years ago

1. 2x-y=1 => y=2x-1 ... sub into equation 1 and solve for x ... 2x-5(2x-1)=-1 => 2x-10x+5=-1 => 10x-2x=5+1 => 8x=6 => x=(8/6)=(4/3) Sub x into either equation and solve for y ... 2x-5y=-1 => 2(4/3) - 5y = -1 => 2.7 - 5y = -1 => 2.7 + 1 = 5y => 3.7 = 5y =>...

9 years ago

Based upon the data given, this problem is indeterminate. One must have the non-standard cell potential for each cell in order to determine the cation concentrations of interest. Without the non-standard cell potentials, there will be two unknowns in each...

9 years ago

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 Liter Solution)...

9 years ago

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 = 12.7

9 years ago

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 + moles N₂)] = 3/(3...

9 years ago

When did Carbon get 7 protons?

9 years ago

[NaOH] = [(25g)/(40g/mole)]/(0.400Liters) = 1.56M

9 years ago

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 problem...

9 years ago

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 = - 0.0.76v (oxidation half-rxn)...

9 years ago

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 anode. Reduction =>...

9 years ago

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 = [(0.008314Kj/mol·K)·ln(2.4 x 10¯² min¯¹/1.1 x 10¯ˢmin¯¹)·(...

9 years ago

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

9 years ago

did this work for you?

9 years ago

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

9 years ago

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 th...

9 years ago

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] =...

9 years ago

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 adds somethin...

9 years ago

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) =...

9 years ago

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. Incr...

9 years ago

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...

9 years ago

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.

9 years ago

How much OH in pH=5 and How much OH in pH=6.5. Determine difference and that's how much CaO is needed.

9 years ago

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.

9 years ago

No one is going to answer a set of MC questions for you.

9 years ago

Not! To much and in wrong units.

9 years ago

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 wou...

9 years ago

1 mole => 6.02x10^23 molecules X => 7.50x10^24 molecules

9 years ago

----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...

9 years ago

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 = (0.00313)(3x)...

9 years ago

Correction ... The [Cr^+3] is 0.039M & the [IO3^-] = 3x = 3(0.039M) = 0.117M. Sorry bout that.

9 years ago

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 = (98g/mo...

9 years ago

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 eleva...

9 years ago

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

9 years ago

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...

9 years ago

Thank you BP for the link. I'm passing it on to my students. Good Job!

9 years ago

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

9 years ago

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...

9 years ago

∆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 work for system < O =>...

9 years ago

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

9 years ago

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 = 0.00...

9 years ago

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.

9 years ago

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 Io...

9 years ago

See your 8:14pm post.

9 years ago

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 ki...

9 years ago

(8 x 7.13)/(2/108) = ________ Kj/mol exothermic

9 years ago

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 + (9.64mol/L...

9 years ago

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)

9 years ago

Also, check assay label for purity. If less than 100% pure, divide grams by fractional purity.

9 years ago

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 con...

9 years ago