A50.0 ml sample of an acid, ha, of unknown molarity is titrated, and the ph of the resulting solution is measured with a ph meter and graphed as a function of the volume of 0.100 m naoh added. at which point on the titration curve is [a-] closest to twice that of [ha] ?

Ch4 and ph3 are both covalent since they are composed both of non-metals.    ch4 is non- polar because of  london forces and has a geometry of tetrahedral  ph3 , on the other hand, is umbrella shaped polar and has dipole-dipole intermolecular force (imf) . since ph3 is polar, it has a stronger imf   

The reaction 2kalsi3o8 + 2(h+ + hco3-) + h2o  →  al2si2o5(oh)4 + 2k+ + 2hco3- + 4sio2 shows that feldspar kalsi3o8 is chemically altered by hydrolysis to form chalky-looking clay mineral kaolinite al2si2o5(oh)4. the chemical breakdown of potassium feldspar due to exposure to acidic water results to the mineral binding with the water molecule to become kaolinite with silica and potassium and bicarbonate ions in solution.

Given data,temperature, t = 230°c = 503 kconsider the chemical reaction,ch₃oh(l) + 1/2 o₂(g)  → ch₂o(g) + h₂o(l)calculate the enthalpy change for the reaction: δh⁰rxn =  σnδh⁰f(products) -  σmδh⁰f(reactants)substitute the values from the table: δh⁰rxn = { -115.9 + (-285.8)} - {0+(-238.7)}              = -163.0 kjcalculate the entropy change for the reaction.δs⁰rxn = (219 + 70) - (127.2 + 1/2 x 205.2) = 59.2 j/kconsider the expression for free energy: δg⁰rxn =  δh⁰rxn - t  δs⁰rxnsubstitute the values in the above equationδg⁰rxn = -163 x 10³ - (503 x 59.2)              = -192777.6 j              = -192.777 kj