Exercise 10: Acid-Base Balance: Activity 3: Renal Responses to Respiratory Acidosis and Respiratory Alkalosis Lab Report Pre-lab Quiz Results You have not completed the Pre-lab Quiz. 07/18/13 page 1 Experiment Results Predict Question: Predict Question 1: What effect do you think lowering the PCO2 will have on [H+ ] and [HCO3- ] in the urine? Your answer : d. [H+ ] will decrease and [HCO3- ] will increase. Predict Question 2: What effect do you think raising the PCO2 will have on [H+ ] and [HCO3- ] in the urine? Your answer : c. [H+ ] will increase and [HCO3- ] will decrease.
2-propanol (bp=82 degrees C) 3. tetrahydofuran (bp=65 degrees C) 4. 1-butanol (bp=118 degrees C) 5. butanone (bp=80 degrees C) Give a better separation for the mixture to be distilled tetrahydofuran (bp=65 degrees C) because it is farthest from 100 degrees C Which alkyl halide would react fastest in a nucleophilic substitution using silver nitrate in ethanol (weak nucleophile, protic solvent)? 3-bromo-3-methylpentane (most
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.
a) for a we first need to find a balanced equation for when the hydrocarbons combust to form CO2 and H20. Then we plug in the deltaHf values and plug these into the equation. a) C4H6 + 11/2O2 ==> 4CO2 + 3H2O Delta Hrxn = [4DeltaHf(CO2)+3DeltaHf(H2O)] - [DeltaHf(C4H6) + 11/2DeltaHf(O2)] = [4(-393.5kJ) + 3(-285.83kJ)] - [111.9kJ + 11/2(0kJ)] = -2543.39kJ C4H8 + 6O2 ==> 4CO2 + 4H2O Delta H rxn = [4DeltaHf(CO2) + 4DeltaHf(H2O)] - [DeltaHf(C4H8) + 6DeltaHf(O2)] = [4(-393.5kJ) + 4(-285.83kJ)] - [1.2kJ +6(0kJ)] = -2718.52kJ C4H10 +13/2O2 ==> 4CO2 +5H2O DeltaHrxn = [4DeltaHf(CO2) + 5DeltaHf(H2O)] - [DeltaHf(C4H10) + 13/2DeltaHf(O2)] = [4(-393.5kJ) + 5(-285.83kJ)] - [-124.7kJ +
C. Write a balanced equation for the neutralization of acetic acid with NaOH. HC2H302+NaOHNaC2H302+H20 D. How would your results have differed if the tip of the burette was not filled with sodiumhydroxide before the initial volume reading was recorded? It would have shown more NaOH then there actually was resulting in there
Name : Amy Vo Date: 11/11/2014 The purpose of this experiment is to determine the equivalence point of the solution. When the conductivity is placed in a solution of Ba(OH)2, the Ba(OH)2 will dissociate into Ba2+ and OH-. The availability of these ions will increase electrical conductivity. When adding H2SO4, the solution will neutralize the Ba(OH)2 producing BaSO4 which is not soluble in water. Therefore, the conductivity will go down.
25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required. Calculate the value of x given the equation: Na2CO3 + 2HCl → 2NaCl + CO2 + H2O 6. 25 cm3 of a sample of vinegar (CH3COOH) was pipetted into a volumetric flask and the volume was made up to 250 cm3. This solution was placed in a burette and 13.9 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Calculate the molarity of the original vinegar solution and its concentration in gdm-3, given that it reacts with NaOH in a 1:1 ratio.
We found the mass of our pure alcohol to be 10.23g. The volume of our pure alcohol was 10.97mL. With these numbers, we used the formula mass/volume to find the density of our pure alcohol to be .9325g/mL. The percent error of our experiment was found to be 18.19% with an absolute error of .1435%. Conclusion:The purpose of this lab is to prepare and purify a fuel, ethanol C2H5OH.
Let’s call this number X. Then the structural formula of BaCl2 hydrate can be written as BaCl2•XH2O. The reaction of dehydration is |BaCl2•XH2O ( BaCl2 + XH2O |(5) | According to the stoichiometry of the reaction (5) |[pic] |(6) | Where N1 is the number of moles of BaCl2 formed in the reaction (5) and N2 is the number of moles of water lost in the reaction (5). To find out N1 you need to divide the mass of BaCl2 after the reaction by its molar mass. To find out N2 you have to determine the mass of water produced in the reaction.
57 (1) (iii) A triglyceride is also formed in the reaction between glycerol and three molecules of oleic acid, C17H33COOH. State and explain which of the two triglycerides (the one formed from stearic acid or the one formed from oleic acid) has the higher melting point. (the one from) stearic acid; saturated / no (C to C) double bonds; chains pack close together / stronger intermolecular forces / van der Waals’ forces etc; 3 Ignore hydrogen bonding. If wrong choice made, only third mark can be scored. (3) (b) An oil sample containing 0.0100 mol of oil was found to react with 7.61 g of iodine, I2.