Olereb48

Oxygen is the limiting reagent in the problem. 320g molecular oxygen = 10 mole oxy. solve (13/10)moles oxy = (8/x)moles CO2. x = 6.15 moles CO2 x 44g/mole CO2 = 270.8 gms CO2

9 years ago

(Molarity x Volume)concentrate = (Molarity x Volume)diluted soln (1M-HCl)(x) = (0.10M-HCl)(100-ml) Solve for x = volume of concentrate that must be drawn and diluted up to but not to exceed 100-ml = 10-ml of concentrate. Transfer to mixing vessel and add...

9 years ago

You are right DrBob, the maximum solubility of CaF2 at 25-deg C = 2.04 x 10^-3M. The problem data is ~2,500X more soluble than what can be achieved.

9 years ago

Amine derivative Weak Bases do not give 1:2 ionization ratios. Balance your equation, then try your problem again.

9 years ago

Better to use ml of concentrate to be diluted to specified volume. Vol Conc(ml) = [(Molarity Needed)(Volume of Soln in Liters)(formula wt of solute)]/[(Decimal Fraction of Solute in concentrate)(Density of conc soln)] Let's assume you need 250 ml of 0.20M...

9 years ago

4Fe+2 + 2SiO-4 => Fe4(SiO4-4)2 However, when writing formulas for Ionic Compounds, reduce the subscripts to the smallest whole number ratio. Therefore, Iron(II) Silicate => Fe2SiO4.

9 years ago

[H+] fm Potassium Sulfate is an hydrolysis calculation using (15g/150ml) = 0.575M sulfate ion in 150 ml soln. Calculate equivalence point [H+](K2SO4) = 1.8 x 10^-8M [H+] fm 0.20M Bisulfate ion is a weak acid in solution calculation in which [H+] is determ...

9 years ago

Follow the sequence: % => gms => moles => reduce moles (divide by smallest mole value) => Emp. Ratio => C:H:N:O C: 63.56% => 63.56g/12g/mol = 5.297mol H: 6.00% => 6.00g/ 1g/mol = 6.000mol N: 9.27% => 9.27g/14g/mol = 0.662mol O: 21.17% => 21.17g/16g/mol =...

9 years ago

For N2(g) + O2(2) => 2NO(g) Each of 2 nitrogen atoms lose 2 electrons while each of 2 oxygen atoms gain 2 electrons. Total electrons transferred = 4 moles electrons. N2 => 2(2e^- lost) = 4 e^- oxidation O2 => 2(2e^- gain) = 4 e^- reduction

9 years ago

This is a dilution problem... Use the dilution equation. (Molarity)(Volume)of concentrate = (Molarity)(Volume) of diluted soln. For these type problems you need to ask 'How many ml of the concentrate do I need to dilute to the needed concentration and vol...

9 years ago

ppm is a weight/weight ratio problem. That is, weight of solute/weight of solution... I am assuming the 0.60ug is in water having a density of 1g/ml. (Sounds like you are running the FeSCN+2 Kc experiment). Anyways, given 0.60ug/ml = 0.6ug solute/1.0g sol...

9 years ago

Σ q = 0 (Mc∆T)water + (M∆Hfusion)ice + (Mc∆T)ice water = 0 [(430g)(1cal/g-oC)(0oC – 87.5oC)] + [(M-ice)(80cal/g)] + [(M-ice water)(1cal/goC)(0oC - 0oC)] = 0 [(430)(-87.5)] + [(M-ice)(80)] + [0] = 0 [(M-ice)(80)] = 37625 => M-ice = (37625/80)g-ice = 470g-i...

9 years ago

What is RTP? STP maybe? Please show what you have done.

9 years ago

1 calorie = 4.184 joules. Basic dimensional analysis. In the 1st conversion, is Calorie referring to a nutritional calorie? If so, 1000 cal = 1 nutritional calorie. Also, basic dimensional analysis.

9 years ago

What have you done on this problem?

9 years ago

Ionize all compounds except iron(III) sulfide which is a ppt driving force.

9 years ago

NaOH + KHPh => NaKPh + HOH Need (2/204.22)mole NaOH = 0.0098-mole NaOH b/c Rxn is 1:1 Rxn Ratio. Ask what volume of 0.50M NaOH would deliver (2/204.22)mole of NaOH to neutralize 2g KHPh or 0.0098 mole KHPh? That is, (0.50M NaOH)(Volume of NaOH soln)= 0.00...

9 years ago

TTL Pressure = 92.3KPa = (692mmHg)* = P(O2) + P(HOH) From reference: P(HOH)=14mm @ 17d-Celcius Therefore P(O2) = P(TTL) - P(HOH) = (692mmHg) - (14mmHg) = 678mmHg. ______________ *P(TTL) = 92.3KPa x (1000Pa/KPa) x (1Atm/101,325Pa) x (760mm/Atm) = 692mmHg T...

9 years ago

Assuming the 71.42gms of limestone is pure CaCO3 then, (71.42g CaCO3/100gms/mole)=0.714mole CaCO3. From CaCO3 => CaO + CO2,the theoretical yield is a 1:1 rxn ratio. Therefore, from 0.714 mole CaCO3, the reaction theoretically delivers 0.714mole CaO and 0....

9 years ago

Work from mole:mole ratios. That is, convert given grams of NO to moles and apply to equation. Let x = moles of N2 produced and set up a linear ratio;i.e, From the equation, 2NO => N2 + 2O2 => (2 mole NO/1.153 mole NO) = (1 mole N2/x) Solve for 'x'= (1/2)...

9 years ago

How about ... P1 = 808 Torr P2 730 Torr V1= 5.65 L Liters V2 ???? liters T1= 17-C Celcius T2 80 OC 273 Kelvin 273 K 290 Kelvin 353 K Pressure Temperature V2 = V1 (P1/P2) (T1/T2) V2 = 5.65-L (808/730) (353/290) = 7.6-L

9 years ago

Also, because the reaction is a 1:1 rxn ratio, it is useful to apply (Molarity x Volume)Acid = (Molarity x Volume)Base => (1.918M)(15.71ml) = M(base) x (26.8ml). Solve for Molarity of base (RbOH). The volumes can be left in ml as both sides of the equatio...

9 years ago

After dissolving the oxalic acid into water, the problem reduces to 1mole Oxalic Acid + 2moles Sodium Hydroxide => 1mole Sodium Oxalate + 2moles Water. In this, moles of NaOH used > moles of OxA neutralized by 2-times. To equate moles of OxA and NaOH => 2...

9 years ago

Mr Pursley is 100% correct, but I would like to add to this in the following way... (that is, in case this question is dealing with practical laboratory preps of solns.) gms NaCl for .9% Soln = (250ml solution)[(0.9g solute / 100g solution)]=2.25 gms NaCl...

9 years ago

Given O3Cl- Al - ClO3 l ClO3 Formal Charge = [Valence - (Bonded Electrons/2) - Non-bonded electrons] FC(Al) = 3 - (6/2) - 0 = 3 - 3 = 0 FC(O) = 6 - (2/2) - 6 = -1 FC(Cl) = 7 - (8/2) - 0 = +3 Net Charge = 1Al + 3Cl = 9O = 1(0) + 3(+3) + 9(-1) = 0 + 9 - 9 =...

9 years ago

Ian, I think you mean PbCl2 will ppt, while Na(NO3)2 remains in aqueous soln.

9 years ago

Let's try this... Adding 0.42 mole of Cd^+2 salt into a 2.50M solution of Cyanide salt. The concentration of Cd(CN)4^2- with a Kf = 7.1 x 10^16 will be 0.42mole/L. (100% conversion) Now, the Kd for the Cd-complex will be (7.1 x 10^16)^-1 = 1.4 x 10^-17. C...

9 years ago

Use ideal gas law. PV=nRT.Solve for n and substitute R=0.08206 L-Atm/mol-K, T=(273+42)K=315K, V=5.2 L, P = 1.00 atm.

9 years ago

Same as last question. Ideal Gas Law. Be sure all data units match units of R-value => L-Atm/mol-K. Substitute and solve for Volume.

9 years ago

Could you please recheck your given CO2, H20 & N2 gram weights. From the values given, the sum of %C + %H + %N from original sample > 100%. Which means %Oxy is a negative number. Not workable for empirical formula analysis.

9 years ago

This is hydrolysis of 'salt' ions... After titration to equivalence point, only salt remains in solution. Before hydrolysis, only salt ions will be present. Using NH4Cl => NH4+ + Cl- only the NH4+ will under go hydrolysis because theoretically Cl- + HOH =...

9 years ago

Use Boyles Law if mass of gas and temperature are unchanged. (P1)(V1) = (P2)(V2). However, you must give temperature change if that is to be taken into account. On the average, temperature of the atmosphere drops 2-deg Celcius per 1000-ft of elevation. Ju...

9 years ago

Based on data given and using a formula wt of CaF2 = 78g/mol; Molar solubility CaF2 = (7E-2/78)M = 8.9E-4M. Ksp = [Ca][F]^2 =(8.9E-4)(8.9E-4)^2 = 7E-10 which does not match published Ksp-values for CaF2. Never the less, In presence of 0.50M CaCl2... CaF2...

9 years ago

Dr Bob is 100% right. You must take the NH3 into account in the formation of the silver-amine complex. It is easier to see if simultaneous equations are written. (1) AgCl <=> Ag^+ + Cl^- (2) Ag+ + 2NH3 <=> Ag[NH3]2^+ _______________________ (3) AgCl + 2NH...

9 years ago

CAUTION!! If working to dilute manufactured concentrate of any acid, be sure to add HCl concentrate to ~ half of final volume needed for solution. Adding water onto 146gms of HCl (~172-ml of concentrated acid) will ionize so fast and give off sufficient h...

9 years ago

Short cut... ________ [H+]= √Ka[Acid] _____________ = √1.75E-5[0.750] M = 0.0036M %Ionization = ([H+]/[HOAc])100% =[(0.0036)/(0.750)]100% = 0.48% is correct for pure water, but NaOAc provides a common ion that reduces the %ionization. Use the Ice table HO...

9 years ago

Assume pressure is constant. It will cancel out of the combined law expression.

9 years ago

q = 80cal/gm x 70gms = 5600cals x 4.184J/cal = 23,430 joules ~ 23,000 joules with 2 sig gigs.

9 years ago

Molarity of Acid x Volume of acid = Molarity of Base x Volume of base

9 years ago

From equation 1 mole KClO3 => 3/2 mole O2 at STP x 22.4L/mole = 33.6L at STP. Given 13.5g KClO3/(122g/mol)=> 3.73L O2 at STP. Adjusting for 40-deg C and 85.4Kpa (~641mm) ... Vol@40C&641mm = 3.73(760/641)(313/273)L = 5.07 Liters

9 years ago

If you can calculate moles O2 ~0.167mole O2 from 13.5g KClO2 and use 313K & (641/760)atm then, PV=nRT is a shorter route.

9 years ago

H2PO4 Undergoes 3 separate ionization steps. Each is treated as a monoprotic ionization. The concentrations at equilibrium for the products of the 1st ionization step are the beginning concentrations for the 2nd ionization and the products of the 2nd ioni...

9 years ago

The 'Drop x' simplification can be determined from this ... if the Conc/Ka > 100 the inherent error will be insignificant. For the 1st ionization I used just (0.10/x10^3) = 100. The concentrations for the 2nd and 3rd ionizations drop off so fast and make...

9 years ago

∏ = MRT = [moles/Vol(Liters)]RT => moles = (∏·Vol(Liters)/RT) = mass(gms)/Mol Wt Solve for Mol Wt = [mass(gms)·R·T/∏·Vol(Liters)] Mass = 4.5-gms R = (0.08206 L-Atm/mole-K) T = (30 + 273)K = 303K ∏ = (0.337/760)Atm = 4.4 x 10^-4 Atm Vol(Liters) = 1.00 Lite...

9 years ago

Are you saying that the coefficients of the balanced reaction equation define order of reaction?

9 years ago

Ms Nicole, you need to re-read the problem and check that the [Cl2] from Exp-1 to Exp-3 was 'doubled' NOT cut in half. The rate decreased to 1/2 the original rate only when the concentration of Cl2 is DOUBLED. If [Cl2] is cut in half and the rate also dro...

9 years ago

BaCl2 + Na2SO4 => 2NaCl + BaSO4 Moles BaCl2 = Moles Na2SO4 Moles = Molarity x Volume(liters) Moles BaCl2 = (3.44gms/207g/mol)= 0.0166 mole BaCl2 which will use the same number of moles Na2SO4. Therefore, Moles Na2SO4 used = 0.0166 mole = (0.282M Na2SO4)(V...

9 years ago

Don't for get about Pressure-Volume effects. Changes in Pressure-Volume will not affect this reaction b/c molar volumes of gas are equal on both sides of equation. However, Vm(Reactants)>Vm(Products), increasing pressure (decreasing volume) shifts reactio...

9 years ago

0.25M H3O^+/(10^-5)mol H3O^+/Liter = 25,000 Liters (one step)

9 years ago

Concentrations that affect electrochemical galvanic potentials are defined in terms of concentrations of ions undergoing oxidation and reduction in their respective cells. That is, when a battery is new, the concentration of ions in solution at the anode...

9 years ago

pH=5.41 after adding 20ml of 0.100M NaOH into 100ml of 0.100M HA

9 years ago

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.

9 years ago

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 0.1...

9 years ago

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 +...

9 years ago

You are welcome.

9 years ago

OK... (1) [H3O^+] = 10^-pH = 10^-5 mol/Liter (2) 0.25moles(H3O^+)/(10^-5)mol H3O^+/Liter = 25,000 Liters

9 years ago

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

9 years ago

You are welcome.

9 years ago

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

9 years ago

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 qu...

9 years ago

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...

9 years ago

You need to specify a volume... The smallest quantity per Liter is given = 3E-25 gram.

9 years ago

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 o...

9 years ago

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 so...

9 years ago

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.

9 years ago

Acronym ... O = Oxidation I = is L = Loss (of electrons) A => A^+ + e^- R = Reduction I = is G = Gain (of electrons) B + e^- => B^-

9 years ago

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, th...

9 years ago

You are most welcome. Doc (aka Olereb48)

9 years ago

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 ca...

9 years ago

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...

9 years ago

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^-...

9 years ago

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.

9 years ago

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...

9 years ago

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.

9 years ago

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 el...

9 years ago

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, HClO...

9 years ago

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

9 years ago

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'...

9 years ago

I got 3600 cals = 15,062 joules = 15.1 Kj

9 years ago

29.5in = 29.5in(25.4mm/in)(0.001316Atm/mm) = 0.986 Atm

9 years ago

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)...

9 years ago

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.,...

9 years ago

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

9 years ago

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

9 years ago

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

9 years ago

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.

9 years ago

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.

9 years ago

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.

9 years ago

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)

9 years ago

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 in...

9 years ago

What if it's a zero order reaction?

9 years ago

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] ......

9 years ago

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...

9 years ago

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 so...

9 years ago

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

9 years ago

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

9 years ago

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...

9 years ago

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!

9 years ago

10g CaCO3 = (10/100)mole = 0.10 mole CaCO3 CaCO3 + Excess HCl =>CaCl2 + CO2 + H2O 0.10-mol CaCO3 + <HCL> => 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

9 years ago

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

9 years ago