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Olereb48
Questions and answers by
Olereb48
Answers (100)
From Ideal Gas Law PV=nRT=(mass/fwt)RT => (mass/Vol)=Density=P(fwt)/RT P = 5 Atm fwt = 16 g/mol R = 0.08206 L-Atm/mol-K T = (273+20)K = 293K Density = [(5)(16)/(0.08206)(293)]g/L = 3.33 g/L
10g CaCO3 = (10/100)mole = 0.10 mole CaCO3 CaCO3 + Excess HCl =>CaCl2 + CO2 + H2O 0.10-mol CaCO3 + => 0.10-mol CaCl2 + 0.10-mol CO2 + 0.10-mol H2O => 0.10(111)g CaCl2 + 0.10(44)g CO2 + 0.10(18)g H2O => 11.1g CaCl2 + 4.4g CO2 + 1.8g H2O
Oh, I see ... 12 moles of any gas will have 6 times more volume than 2 moles of any gas at equivalent conditions. In this case STP. Ha! 1st read was really confusing, but I get it!
Aluminum will not react with NaOH. Sodium (Na) is higher on the activity series than Aluminum. If you are referring to Al + 3HCl => AlCl3 + (3/2)H2 then you have a reaction that will produce hydrogen gas. Assuming rewrite ... How many grams of Al will be
Vol of each = (22.4L/mol)(12 moles) = 268.8 Liters Both have the same volume. 1 mole of 'any' gas at STP occupies 22.4 Liters. So, 12 moles of 'any' gas (including CO2(g) and CO(g) at STP occupy 12(22.4)Liters = 268.8 Liters
Rate Law with both reactants 1st order: Rate = (0.0027L/mol-s)[S208^2-][I^-] [S2O8^2-] = (0.006L)(0.10M)/(0.010L) = 0.06M [I^-] = (0.004L)(0.20M)/(0.010L) = 0.08M Rate = (0.0027L/mol-s)(0.06mol/L)(0.08mol/L) = 1.3x10^-5 mol/L-s
BeCl2 has a linear crystalline molecular geometry with bond angle of 180-deg. X-A-X geometry by the VSEPR Theory and SiO4 is a tetrahedral molecular geometry with 104.5-degree bond angles with an AX4 geometry. SiO4 can not form closely packed molecular
This is a common ion problem where 0.10M NaOH is added into a solution containing 0.0001M Ca(OH)2. Ca(OH)2 Ca^+2 + 2OH^- Ci: -------- 0M ---- 0.10M ∆Ci: ------- +x ---- +2x Ceq: -------- x ---- 0.10+2x~0.10M Ksp = [Ca^+2][OH^-]^2 6.5x10^-6 =
This is a common ion problem where 0.10M NaOH is added into a solution containing 0.0001M Ca(OH)2. Ca(OH)2 Ca^+2 + 2OH^- Ci: --- 0M 0.10M ∆Ci: --- +x +2x Ceq: --- x 0.10+2x~0.10M Ksp = [Ca^+2][OH^-]^2 6.5x10^-6 = [Ca^+2](0.10)^2 Solve for [Ca^+2] ... =
What if it's a zero order reaction?
To answer this question you need to understand the concept of system vs surroundings and how the terms endothermic and exothermic are related to this and applied to define heat flow associated with a reaction process. The system is always the object of
Work from the definition of specific heat... The amount of heat needed to raise 1 gram of substance 1 deg C => cal = 1 gram per deg C => Specific Heat = cal/(g/deg C) = cal/(g-deg C)
From the 'balanced equation', 2 moles Cl2O7 requires 130 kcals => normalizes to 1 mole Cl2O7 requires 65 kcals. From this 5 moles Cl2O7 then requires ... ??? You do the math.
Transition state stage. The greatest probability for reactants to proceed to products is when reacting components have sufficient energy to cross the 'activation energy' point of the reaction's potential energy diagram.
Increasing pressure shifts to the lower molar volume side ... Reactant Side has 5 molar volumes of gas and the product side has 4 molar volumes of gas. Increasing pressure would shift reaction to the 4 molar volume side; that is, shifts rxn right.
For a 1:1 rxn ratio of monoprotic acid reacting with a GpIA strong base use ... (Molarity x Volume)acid = (Molarity x Volume)base Molarity Acid = [(M x V)base]/(Vol Acid) = [(0.5M)(45.5 ml)]/(150 ml) = 0.16M HOAc
Gms of HOAc = (35.0ml)(1.049g/ml) = 36.715gms HOAc moles HOAc = (36.715 g)/(60.05 g/mol) = 0.6114 mole Molarity = moles/Liters = (0.6114-mole/0.400-L) =1.53-Molar
Using Ideal Gas Law PV = nRT = (mass/f.wt)RT Solve for f.wt. f.wt = (mass)RT/PV = (2.45g)(0.08206L-Atm/mol-K)(418K)/(1-Atm)(1-L) = 84.04 g/mole
A: 2d & 1f do not exist B: 3f & 1p do not exist The 1st number is the principle energy level (ring number) and the letter is the orbital shape. The max number of orbitals per energy level is defined by the value of n. That is, n=1 has only 1 orbital;i.e.,
From KM-Theory => RMS velocity = 158(Sqr-Rt(T/M)) meters/sec F.Wt(Ar) = 40 g/mol F.Wt(Rn) ~ 222 g/mol RMS V(Ar) = (158)[Sqr-Rt(400/40)] m/s ---------> = 500 m/s RMS V(Rn) =0.5(500) m/s = 250 m/s Solve RMS Velocity Equation for 'T' => (250/158)^2 = (T/222)
29.5in = 29.5in(25.4mm/in)(0.001316Atm/mm) = 0.986 Atm
I got 3600 cals = 15,062 joules = 15.1 Kj
Another way is through the KM-Theory where RMS velocity = 158(Sqr-Rt(T/M)) meters/sec F.Wt(Ar) = 40 g/mol F.Wt(Rn) ~ 222 g/mol RMS V(Ar) = (158)[Sqr-Rt(400/40)] m/s ---------> = 500 m/s RMS V(Rn) =0.5(500) m/s = 250 m/s Solve RMS Velocity Equation for 'T'
14.7lbs/in^2 = (14.7lbs/in^2)(454g/lb)(1x10^-3Kg/g)(0.155in^2/cm^2)(1x10^4cm^2/m^2) = 1.03x10^4kg/m2
Electrolyte ionizations fall into 2 categories ( 3 if you include non-electrolytes )... Strong Electrolytes => Ionize 100% Weak Electrolytes => Ionize < 100% Non-Electrolytes => Ionize 0% Strong Electrolytes include: Strong Acids (HCl, HBr, HI, HNO3, HClO4
This is an electrolytic cell used for (most likely) gold plating. The electrode connected to the positive terminal of the battery will be positive and attract anions from a gold salt rich solution to the electrode to undergo oxidation (=> cathode). The
Rx shows a 2 to 12 Hexane to CO2 ratio, or (if reduced) a 1 to 6 mole Hexane:CO2 ratio; that is, for each mole of hexane burned, 6 moles of CO2 are produced. So, if given 18.4 moles Hexane then... 6 times more in moles of CO2 will be formed. = 110 moles.
This is a 1:2 ionization in pure water => Solubility of any 1:2 or 2:1 salt ionization = Cube Root(Ksp/4). Solubility (pure water) of Mg(OH)2 = Cube Root[(9x10^-12)/4] = 1.3x10^4 M. 1:1 ionization => S = Sqr Rt(Ksp) 1:2 or 2:1 => S = Cube Rt(Ksp/4) 1:3 or
Na2CO3 => 2(Na^+) + 1(CO3^-2) The CO3^-2 is the common ion and addition of more CO3^-2 will overload the PRODUCT side causing the equilibrium to shift left to remove the excess and re-establish a new equilibrium.
845ml(2.5x10^-5M Ni(NO3)2 + 195ml(0.0486M KOH) =>0.845(2.5x10^-5)mole Ni(NO3)2 + 195(0.0486)mole KOH =>(2.113x10^-5)mole Ni(NO3)2 + (9.48x10^-3)mole KOH =>(2.113x10^-5)mole Ni^2 + (9.48x10^-3)mole OH^- =>(2.113x10^-5 mole/1.04 Liter Soln)Ni^2 + (9.48x10^-3
Since from the balanced equation, both NO and NO2 have the same coefficients, then 5 moles of NO will produce 5 moles NO2. Convert moles to grams by multiplying formula weight of NO2 by 5 moles, giving grams of NO2 produced. grams of NO2 = (46 gms/mole)(5
The Cobalt Chloride-Hexahydrate I remember is more purple than pink. It is easily dehydrated into blue crystals. The anhydrate is highly hydroscopic (adsorbs water) returning to purple (pink) color. In the anhydrate form, appropriately sized quantities can
You are most welcome. Doc (aka Olereb48)
Are you familiar with the simplification concept when calculating hydronium ion concentrations or weak acids in pure water? Basically, it says that in the I.C.E. Table, at equilibrium, if the concentration of weak acid divided by the Ka value is > 100,
Acronym ... O = Oxidation I = is L = Loss (of electrons) A => A^+ + e^- R = Reduction I = is G = Gain (of electrons) B + e^- => B^-
Correction... Molality(m) remains the same if temp is changed... (next to last sentence). remove the 'not'. If there's an edit feature on this site, I don't know how to use it. sorry bout that.
Molality(m) = moles solute/Kilograms Solvent For a 10.0% w/w aqueous solution of EtOH(f.wt. = 46g/mol) => 10gms EtOH/100gms Soln. Using this ratio as solute-in-solution base solution, then the data needed for molality would be 10gms EtOH in (100-90)gms
The symbology should be lower case 'p' followed by capital 'H' => pH ... This notation comes from p-factor analysis or pX, and 'X'designates quantity of interest. pH is only one of several p-factor conversions. In the case of pH, it is used for analysis of
You need to specify a volume... The smallest quantity per Liter is given = 3E-25 gram.
If a sodium chloride aqueous solution is electrolyzed with a selected current (10 amps in this case), only Water will undergo oxidation and reduction and pH will = 7. Also, neither Na^+ nor Cl^- will hydrolyze in water and the pH will be exclusively upon
All are sp^3 hybrid derivatives of tetrahedral parent structure. 'pi' bonds are found in double and triple bond system... Pi bond electrons are (by 'textbook definition') the sideways overlap of unhybridized p-orbitals. So, none of the listed compounds
Given 0.84M HOAc and Ka =1.8x10^5 For a weak monoprotic acid the [H] can be calculated from [H] = (Ka[Acid)^1/2 = [1.75x10^-5(.83)]^1/2 = 3.8x10^-3M(H) pH(vinegar) = -log[H] = -log(3.8E-3) pH = 2.42
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RMS = [158(Kelvin Temp/Mol Wt)^1/2]meters/sec RMS O2 = 158(300/32)^1/2 m/sec = 484 m/sec Kelvin Temp for N2 RMS to equal O2 RMS @300K = M(RMS/158)^2 =28(484/158)^2 = 263 Kelvin Check ... RMS N2 = 158(263/28)^1/2 m/sec =484 m/sec
OK... (1) [H3O^+] = 10^-pH = 10^-5 mol/Liter (2) 0.25moles(H3O^+)/(10^-5)mol H3O^+/Liter = 25,000 Liters
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Given 1gm of MM contains 8% Mg(OH)2 (f.wt. = 58.33 g/mol) Gms Mg(OH)2 = 8% of 1gm MM = 0.08gms Mg(OH)2 = 0.00137mol Mg(OH)2 Mg(OH)2 in solution => 2moles OH^- Moles OH^- = 2(0.00137)mole OH^- = 0.00274mole OH^- Adding 50ml of 0.100M HCl => 0.005mole HCl +
1-g of MM contains 8 wt% Mg(OH)2 (f.wt=58.33g/mole) 8% of 1gm = 0.08(1)g Mg(OH)2 = 0.08g Mg(OH)2 = (0.08/58.33)mole Mg(OH)2 = 0.00137 mole Mg(OH)2 Mg(OH)2 in solution => Mg^+2 + 2OH^- moles of OH- = 2(0.00137)mole OH^- = 0.00274mole OH^ Adding 50ml of
I would have posted my solution, but the site monitor keeps kicking out my post as a website post. This is an acid/base titration with a common ion. Hope you can get a solution.
pH=5.41 after adding 20ml of 0.100M NaOH into 100ml of 0.100M HA
