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Divide 52728 by the smallest number so that the quotient is a perfect cube. Also find the cube root of the quotient
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PHI 208 Week 2 - Readings Quiz
1. Tom Regan’s view of animals is that (Points : 1) They are important but not quite as
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Find three examples of lessons from the front line that are evident in the Nestlé case. How could these issues be overcome?
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Please, what is the balanced equation for Potassium Bitartrate and Sodium Bicarbonate? Is there any easy way to remove the
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What would be some good reasons for using figurative language in writing stories?
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Which of these are considered figurative language? dialect,personification, protagonist, metaphor, simile, ambiguity
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Answers (44)
Great Scott!
Using Standard Heats of Formation and Standard Entropy values at 25C apply to free energy equation => ∆G(f) = ∆H(f) - T∆S ∆H(f) = -220.1 Kj/mole (direct from thermo properties table) ∆S = -47.34 KJ/mole-K (from Hess's Law)* T = 298K *=> C +
Since cell is Cr/Ni, it shouldn't have to say it's spontaneous. The thing that's missing is the standard cell voltages. With that it's spontaneous. Flow is spontaneous from more negative to more positive electrode potential.
Since cell is Cr/Ni, it shouldn't have to say it's spontaneous. The thing that's missing is the standard cell voltages. With that it's spontaneous.
There's only one way a Cr/Ni Rxn can go... Cr(
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) => Ni(
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) => Std Cell = 0.51v For non-Std Cell => [Cr/Ni^+2[0.010M]//Cr^+3[0.875M]/Ni]; n=6 2Cr + 3Ni^+2 => 2Cr^+3 + 3Ni Q = ([Cr^+3]^2/[Ni^+2]^3) Apply to Nernst Equation => E(Non-Std) =
Be sure to check your ion concentrations in the Nernst Equation are raised to the proper power. Cell [Cr/Ni^+2[0.010M]//Cr^+3[0.875M]/Ni] Q = [Cr]^2/[Ni]^3 = (0.875)^2/(0.010)^ = (0.766)/(1.0E-6) = 7.66E+5 E^o = E(Redn) - E(Oxdn) = (-0.23v)-(-0.74v) =
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The !-X is supposed to be over and under the A.
Drawing lewis structures 1. Calculate Valence Number (total #valence electrons - from group number) 2. Calculate Octet Number (total #electrons after bonding - always 8 (nonmetals) except for hydrogen and complex substrates => 2) 3. Calculate Bond Number
Rate Factors => C.A.N.T.C. C => Concentration Effects of Reactants A => Area (Surface Area) N => Nature (Chemical Structure) T => Temperature and Energy Contents C => Catalytic Effects Rxn Rate = f(Rate Factors) => Rate Laws
What was wrong with the solution I posted on this same problem? => Tuesday, April 18, 2017 at 9:35pm
C2H6 + 7/2O2 => 2CO2 + 3HOH moles O2 consumed permole C2H6 = 7/2 mole = 3.5 moles O2 per mole C2H6.
Zn + 2HCl => ZnCl2 + H2 Moles Zn = (4.25g/65.37)= 0.065mole Zn available. Moles HCl = 25ml(3M HCl) = 0.025(3) mole HCl = 0.075 mole HCl HCl is limiting reagent => Only 1/2(0.075 mole)Zn = 0.0375 mole Zn will be consumed in reaction => (0.065 - 0.0375)=
Sulfur combines with H+ => H2S(g) and shifts the solubility equilibrium toward more ionization and increased solubility. The formation of H2S essentially is removing S-2 from the solution forcing the metal-sulfide to decompose in order to replace the
0.60M HF/1.00M KF Buffer Soln Before adding OH- or H+, Ka=[H+][F-]/[HF] => [H+]=Ka[HF]/[F-] => [H+]=4.2E-4(1.00)/(0.60) => [H+]=0.0007M in H+ => pH=-log(0.0007)=3.15 HF => H+ + F- 0.60 x 1.00 Adding 0.08M NaOH Shifts Equilibrium Right, decreasing HF by
Solubility in pure water of all 1:2 or 2:1 ionizations with Ksp values is => Cube-Rt(Ksp/4). For Ca(OH)2 ; Ksp=6.5E-5 Solubility = Cube-Rt(6.5E-5/4) = 0.0118M in Ca^+2 and 2(0.0118M)in OH^-. In pure water (no common ion effect) the following can be used to
4p holds 5 electrons in the ground state with two orbitals having paired electrons and one orbital have an unpaired electron. Br:[Ar]4(s^2)3(d^10)4(px^2)4(py^2)4(pz^1)
Try V(rms)= [158 x SqrRt(T/M)]m/s T = Temp in Kelvin M = Molecular weight
You need to review the Aufbau Diagram. It shows the order of electron entry for the building up process. For Mo, it is in the 5th row of the periodic table => Mo:[Kr]5s^24d^4 ... for Mo(+3), ionization of transition metals always starts from the valence
Convert given data to moles, relate to the reaction ratios, then convert back to grams. a) for 0.12g Ag = 0.0011mole Ag => produces 1/2(0.0011mole) Ag2S => (0.00056mole)(248g/mole) = 0.138g Ag2S b) assume an excess of H2S is present, then the rxn will
Heat flow in problems like this use q = mcT if a temp change is observed, and q = mH if a phase change is occurring. Here m = mass of substance gaining or losing heat, c = specific heat of substance, T = Temp change and H = Transition Constant for phase
Using the Henderson-Hasselblach equation => Salt to Acid Ratio =[NaHCO3]/H2CO3] = 0.238/1. Thus, for a .1M H2CO3 the buffer would be 0.0238M in NaHCO3. => grams of NaHCO3 needed = 0.0238 mole(84 g/mole) = 20 grams. Adding 0.02mole HCl to the buffer
If pH(1) = 1.7 => [H+]=10^-1.7 = 0.020M If pH(2) = 1.7 + 1.4 = 3.1 => [H+]=10^-3.1 = 0.0008 pH(1)/pH(2) = 0.020/0.00080 = 25X dilution If pH(3) = 2.2 => [H+]=10^-2.2 = 0.0063M If pH(4)= 2.2 + 1.4 = 3.6 => [H+]= 0.00025 pH(3)/pH(4)= 0.0063/0.00025 = 25X
[mc∆T]Cu + [mc∆T]HOH = 0 [mc(Tf - Ti)]Cu + [mc(Tf - Ti)]HOH = 0 All data is given in problem except T-final. Solve for Tf. m(Cu) = 33.2 gm m(HOH)= 50.0 gm Ti(Cu) = 70.5 C Ti(HOH)= 25.0 C Tf = ??? c(Cu) = 0.385 J/gC c(HOH)= 4.184 J/gC Your answer should
(1)Ca. Moles H = Molarity x Volume(L) = 0.005L(0.0983M)HCl = 0.00049 mole HCl in 50ml of solution. Now, covert to [HCl] = (moles HCl)/(Vol. Soln in Liters) = 0.00049mole/0.050L =0.0098M in HCl => [H]=0.0098M. (2) pH = -log[H] = -log(0.0098) = 2.01
P(1) = 745mm P(2) = 0.115atm(760mm/atm)= 87.4mm V(1) = 414ml V(2) = ? T(1) = 24C = 297K T(2) = -95C = 178K (PV/T)1 = (PV/T)2 Substitute above data into equation and solve for V(2).
1. Use ideal gas law to determine moles of ProAm. Be sure to convert given data to units of R-Value. 2. Determine concentration in 0.500L soln. =[moles ProAm/Volume(L)]. 3. Determine [OH]=SqrRt([ProAm]Kb) 4. Determine pOH=-log[OH] 5. Determine pH=14-pOH (I
just use v(rms) = (158)[SqrRt(T/M)] meters/sec ... T = Kelvin & M = gms/mole ... 158 is from SqrRt[(3)(8.314)(1000)] ... These are constants factored out of the original equation v(rms) = SqrRt(3RT/M).
Same as in your 2:27pm post
1 mole of any gas occupies 22.4 liters at STP.
correction-2 => 0.0017 mole KClO4! not O2.
Correction... I'm used to decomposing KClO3, should of used f.Wt. of KClO4 = 138g/mole... Same logic applies to all else. Moles KClO4 decomposed = 0.0017mole O2, then => I get 0.235g KClO4 decomposed ... Sorry bout that.
See 6:36pm post by Sam.
Calculate dry wt of Calcium Carbonate without watch glass... Then, ratio [CaCO3(s) / tablet mass] x 100% => Wt% CaCO3 I get 55.4% w/w.
See 10:14pm post.
Use Ideal Gas Law (PV=nRT & n=PV/RT)=> moles of O2... Divide by 2 => moles of KClO3 ... Multiply moles KClO3 by formula wt of KClO3 => grams of KClO3 decomposed. Be sure all data is converted to same units as R-value. I get 0.021 gms KClO3. Good luck.
Faraday's Law... 1 mole electrons = 96,500 amp-sec. => ?gAg = 5amps(1mol e/96,500amp-sec)(1 mole Ag/1 mole e)(107g Ag/1 mole Ag)(60sec/min)(50min) = You punch it into your calculator.
You'll need the average density of whole blood. (~1060 Kg/m^3) take 0.100% of this and convert to grams/L.
1. Write and balance equation 2. Convert given solution data to moles solute ... moles = Molarity x Volume in Liters 3. Apply moles reactants to balanced equation (divide each mole amount by respective coefficient => Limiting Reagent) 4. Using moles of
The reaction is a 3 to 1 ratio Pb(NO3)2 : Al2(SO4)3 => moles aluminum sulfate = 1/3(moles lead nitrate) => 1/3[(0.02549L)(0.1338M)] = 0.00114 mole aluminum sulfate => Therefore, concentration = [Al2(SO4)3] = 0.00114 mole/0.02549L = 0.045M Aluminum Sulfate.
Convert gram values given to moles. (moles = mass / f.wt), then divide the mole values by respective coefficients... the smaller number is the limiting reagent. Moles N2 = 5.42/28.02 = 0.193; Moles H2 = 5.42/2.02 = 2.68 => compare N2(0.193/1) and
1. Question : Which of the following is an impact of increased ocean acidification as caused by global climate change? Student Answer: CORRECT Low pH seawater alters calcium carbonate concentrations which reduce the ability of many animals to build shells.
The answer is 6. The rooms could just have well been labeled A, B, C. 6-4=2 Room A 6-4=2 Room B 6-4=2 Room C 2+2+2=6 or since all but 4 are going into each room, the 4 going into other rooms are split equally into 2 and 2.
2. plasma 6. all of the above rest correct
