how many moles of iodomethanel are in 4dm cube of 2M aqueos solution?
912770
Conversion Factors\n1L=.001 cubic meters, 1 cubic meter =10 dm cube so\nConverting to Liters of solution\n4dm=.4 cubic meters=400L\nUsing ratio to find mols/Liter (which is M)\n2M=2mol/1L=x/400L\n\nx=800mol of iodomethanel in 4dm cube of 2M aqueos solution\n\nnext question\n500cm=5meters cubed=5000L\n1M=1mol/1L=5000mol/5000L\n5000mols of Sodium Hydroxide\n\nThe next question is not a complete question, I'm not getting it.
868152
1 mol NaOH and 40 g NaOH are equivalent.
117753
Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
813561
The molecular mass of NaOH is 40.00 gm/mol,  so 8 gm of NaOH contains 8 gm/40.00 gm.mol = 0.20 moles of NaOH.\n\nThe molecular mass of HCl is 36.46 gm/mol, so 8 gm of HCl contains 8gm/36.46 gm/mol = 0.22 moles of HCl.\n\nIf we react these amounts of NaOH and HCl, the NaOH will be the limiting reactant, and we will produce 0.20 moles each of NaCl and H2O.  This reaction will liberate \n\n0.20 mol * 56 kJ/mol = 11.2 kJ of energy as heat\n\nThe heat capacity of the final solution is given by the mass of the final solution (which, by conservation of mass, is equal to the mass of the initial solutions) multiplied by the specific heat capacity of the final solution.  With the assumptions given in the question, the heat capacity of the final solution is:\n\n(58 gm + 258 gm) * 4.184 J/(gm*degC) = 1322 J/degC.\n\n(Note that we don't need to use the density of the final solution.)\n\nThe temperature change of the solution is given by the energy released divided by the heat capacity of the system:\n\n(11.2*10^3 J)/(1322 J/degC) = 8.47 deg C\n\nIf the initial solutions had a temperature of 25 C, the final solution has a temperature of 25 + 8.47 C = 33.47 C.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
787311
A. H2O. Molecular water is 2H2O
725559
Take, say, a standard 2M solution and dilute it by a factor of 20.\n\nThat is, add 19cm3 of water to 1cm3 of 2M HCl.
354896
1400 mL of NaOH needed...\n\nneed to make sure the concentrations are the same in order for them to completely neutralize\n\nM = Molarity... number of moles of solute in one liter of solution\n\nif you have 3 M for the sodium hydroxide you would need twice that amount in order to neutralize the 6 M of hydrochloric acid\n\nyou would need 1400 mL since the number of moles of the sodium hydroxide is exactly half of that of the hydrochloric acid.  With knowing that the number of moles NaOH is half, you would need double of what you have for the HCL...  700x2 is 1400 mL
308409
Good question and I am the right one to answer it for you.\n\nBy definition\nan iota = 1 molecule\nweight of one molecule of water = 18 moles. (1 mole of H2O = 6.023 X 10^23 molecules of H2O = 18 grams H2O) \nso, we can establish\nan iota of water = 18 grams (1kg of water = 1 litre of water)\n\nA gallon of water in US = 3.785 litres\nA gallon of water in UK = 4.546 litres\nTherefore 5 gallons would contain 1051.38888888888889*6.023* 10^23 iotas in the US\nand in the UK 5 gallons would contain 5051.1111111111111*6.023*10^23 iotas.\nHope your problem is solved.
916567
STP is 1 atm, 273 K.\n\nmoles of HCl = PV/RT = 1 atm * 1.1 L/ (0.08206 * 273K) = 0.049 moles\n\nSince HCl and NaOH is a 1:1 reaction, 1 mole HCl gives 1 mole NaCl, so you get 0.049 moles NaCl (MW 58.5), or 2.87g.
725692
To neutralise 2 moles of NaOH you need 1 mole of H2SO4\nTo neutralise 2 moles of NaOH you need 2 moles of HCl\nUsing the formula: moles = MV/1000 ;M = molarity; V = volume in dm3, then:\n\na) HCl would require twice the volume as H2SO4 to neutralise an 2 moles of NaOH.
834810
You have 75mL of 0,4000M NaOH solution\n0,4000 M= moles of NaOH/0,075L\nmoles of NaOH = 0,030\n 3NaOH + H3PO4 → 3H2O + Na3PO4\nNaOH reacts with H3PO4 at a ratio of 3 moles : 1 mol\nTherefore: 0,030 m NaOH*(1m H3PO4/ 3m NaOH)= 0,010 moles of H3PO4\nThen\nM = 0,010 moles/0,050L\nM = 0,20\nYou are correct; nevertheless, watch the figures!!!!
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
507416
I2 + H2O = IO3(-) + H+\n\nI2 = I-
413725
4 litres\n\nThere are 8 mols of chloride ions in 2 litres of 4M HCl solution. Therefore, the volume of 2M NaCl solution that has the same number of chloride ions is 8mols/(2mols/liter)=4 litres.
808549
M=molesof solute/Liter of solution\n\nM= molarity