An excess of sodium hydroxide is treated with 1.1 L of dry hydrogen chloride gas measured at STP. What is the?
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.
919726
Sodium and Potassium are correct.  They are extremely reactive with water, forming Hydroxides and releasing Hydrogen gas, which is usually ignited from the exothermic heat of the reaction.\n\n2 Na + 2 H2O -> 2 NaOH + H2\n\n2 K + 2H2O -> 2 KOH + H2
792924
A.\n2*22.4 l.\n\nB.\n1*22.4 l.\n\nOne mole of any gas occupy 22.4 l. at STP.
237268
.001565mol / .075L = .021M NaCl
993248
HCI hydrogen chloride \nto make HCI u have to mix Sulfuric Acid and Sodium
113508
If "salt" is Sodium Chloride, then you will be making Hydrogen gas (H2) and (Oxygen gas (O2) and/or Chlorine gas (Cl2)).\n\nUnless you plan on very carefully controlling the potential difference across your electrolytic cell to use just the bare minimum voltage, you are likely to make Chlorine gas at the anode instead of Oxygen.\nOxygen gas has a reduction potential of 1.23 volts, where are Chlorine gas has a reduction potential of 1.36 volts.\nThis means that if you supply just a tiny bit of extra voltage across the cell, instead of just producing Oxygen gas from the water, you will be oxidizing the Chloride ions in solution to make Chlorine gas.
311730
I think you should keep it simple yet smart. First get a piece of pure elemental Sodium and then let it drop on warm water. Afterwards explain how the chemical reaction released Sodium Oxide and Hydrogen Gas.
606155
NaOH + HCl ---> NaCl + H20
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
853370
The concept of 1.5M hydrogen gas is rather odd, but, yes, you use the Nernst Equation.  Don't forget to convert the pH to a molarity of H+ ions.
985020
1 mol = 22.4 litre at STP\nmass = mol x molecular weight\ncause all are 1 mol than all are 22.4 litre\na) 1 mol H2O = 18 gram\nb) 1 mol CH3Cl = 50.5 gram\nc) 1 mol SO4 = 96 gram\nd) 1 mol NH3 = 17 gram\ne) 1 mol C3H7O = 59 gram\nd) 1 mol NO2 = 46 gram
590355
Magnesium chloride (MgCl2) and hydrogen (H2) is the gas released in the reaction.
133426
The density of a sodium hydroxide solution will increase as the concentration of sodium hydroxide in dissoved in water increases.
101093
I don't even get your question here... \n\npH is a mesurment of the quantity of Hydrogen Molecules that is available in a solution.  The pH of a solution changes depending on the concentration of salt you put in.
957626
It is because you have the wrong compound (HSO3)!\n\n2 HI + K2SO3 ---> 2KI + H2SO3
302581
add 2 parts HCl 1 part H20 to dilute properly
780228
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70