Calculate Kp at this temperature for the following decomposition reaction?
A sample of solid ammonium chloride was placed in an evacuated container and then heated so that it decomposed to ammonia gas and hydrogen chloride gas. After heating, the total pressure in the container was found to be 4.4 atm. Calculate Kp at this temperature for the following decomposition reaction. NH4Cl(s) NH3(g) + HCl(g)
Titration calculation question?
Lancenigo di Villorba (TV), Italy SODIUM CARBONATE, e.g. Na2CO3, is a Sodium Salt of an acid like Carbonic Acid (e.g. H2CO3). So, since HYDROCHLORIC ACID (e.g. HCl) RESULTS A STRONGER ACID THAN CARBONIC ACID, HCl PLAYS A DOUBLE DISPLACEMENT WHILE IT MEETS Na2CO3 i) Na2CO3(aq) + HCl(aq) ---> NaHCO3(aq) + NaCl(aq) ii) NaHCO3(aq) + HCl(aq) ---> H2CO3(aq) + NaCl(aq) WHICH HAVE TO MOVE RIGHTWARD SINCE ITS Equilibrium Constant ARE LARGE NUMBERs K,i = 1 / Ka,2 = 1 / 5.6E-11 = 1.7E+10 >> 1 K,ii = 1 / Ka,1 = 1 / 4.6E-7 = 2.3E+6 >> 1 So, YOU STATE THAT ALL Hydrochloric Acid HAS BEEN SPENT TO CONVERT Sodium Carbonate INTO Carbonic Acid. Then, YOU HAVE TO CALCULATE Moles of Hydrochloric Acid moles, HCl = Molarity, HCl * Volume, HCl YOU SPENT UNTIL EQUIVALENCE POINT : YOU HAVE TO BEHALVE THIS MOLAR AMOUNT mol,HCl / 2 = formula-grams,Na2CO3 AND YOU ACHIEVE Formula-Grams OF Na2CO3. NUMERICAL EXAMPLE Some crystals of Na2CO3 has been dissolved into 50 mL of Pure Water. Well, you executed a titrimetric determination by an aqueous 0.100 M HCl : this acidic liquid flown down by a burette and you dropped down 15 mL of it until its Equivalence Point, e.g. Orange Methyl turns reddish and liquid begins to fizz out. I LOOK FOR Formula-grams OF Na2CO3 0.100 * 15E-3 = 1.5E-3 moles of HCl 1.5E-3 / 2 = 7.5E-4 formula-grams of Na2CO3 I hope this helps you.
Calculate the pH of each of the following solutions.?
Part 1- (a) HA <--> H^+1 + A^-1 Ka = 1.3x10^-5 = [H+][A-]/[HA] let X = mol/liter that ionize Ka = X^2/(0.100 - X) assume X << 0.100 1.3x10^-5 = X^2 / 0.100 X = 0.00114 = [H+] pH = 2.94 (b) A- + H2O = HA + OH- Kb = [HA][OH-] / [A-] Let X = mol/liter that hydrolyze Kb = Kw/Ka 1x10^-14 / 1.3x10^-5 = X^2 / (0.100 - X) assume that X << 0.100 X = 8.77x10^-6 = [OH-] so pOH = 5.06 pH = 14.00-5.06 = 8.94 (c) 7.00 (d) when [A-]/[HA] = 1, pH = pKa pKa = -log(1.3x10^-5) = 4.89 = pH Part 2 (a) HCl predominates; pH of 0.03 M HCl = -log(0.03) = 1.52 (b) use the Henderson-Hasselbalch equation; HA = 0.030 M and A- = 0.097 M (c) same as (a) (d) use the Henderson-Hasselbalch equation; HA = 0.130 M and A- = 0.097 M
Le chatelier's principle?
a. If more reactant is added, the equation will shift to the right in order to make more product (which will increase the products) b. If the temperature is decreased, the reaction will shift to the right to create more heat if it is exothermic (which will increase the product), and to the left if it is endothermic (which will decrease the products) c. If the pressure is decreased, it will shift in the direction where there are more moles of gas present, because it can now take up more space. If the products have more moles of gas, the products will increase, and vise versa.
Write the complete balanced equation for the reaction that occurred in this lab. Hint: H2CO3 is not a final pr?
The good rule to work with is that if you react a base with an acid you produce a salt and water. But in if the base is a carbonate, as you have here, then in addition you produce carbon dioxide gas. Put this into a balanced equation: NaHCO3 + HCl → NaCl + CO2 + H2O You reacted 37.0 - 24.35 = 12.65g of NaHCO3 Molar mass NaHCO3 = 84.0g/mol 12.65g = 12.65/84 = 0.151 mol NaHCO3 From the balanced equation, 1mol NaHCO3 will produce 1 mol NaCl 0.151mol NaHCO3 will produce 0.151 mol NaCl Molar mass NaCl = 58.44g/mol 0.151mol = 0.151*58.44 = 8.80g NaCl produced- theoretical yield Actual yield = You claim : salt produced = 31.52g - this is not possible. Could it be: salt produced = 31.52-24.35 = 7.17g ( I subtracted the mass of evaporating dish) This seems more reasonable value % yield = actual yield / theoretiical yield * 100 = 7.17/8.80*100 = 81.5% If you had not driven off all the water, the mass of NaCl produced would be higher than it should be. Your actual yield would be too high, and the % yield would have been higher than the correct value.