Question

please answer as many questions with detail.

1. Evaluate how much "weight" (mass, strictly speaking) an average human loses daily due to gas exchange while breathing. Find all necessary data, indicate your sources, and present a detailed calculation. 2. 10.00 g of a Mg-Al alloy completely dissolved in hydrochloric acid. The volume of hydrogen gas given off in the chemical reaction is 12.29 L Find the mass composition (in mass %) of the alloy. Assume that the volume of 1 mole of gas is 22.4 L. 3. Review the problem on pp. 6-10 of lecture 3 on chapter 3 (see pdf on Moodle) and consider isomers (compounds that have the same molecular formula, but different struc- tures) on p. 10. Based on the elemental composition and molar mass, we cannot tell these substances apart. To do so, additional information is required. Part A. Without consulting any sources, range the substances on p. 10 by their melting or boiling point. Explain your answer. Part B. Compounds with different structures produce different ionic fragments in a mass spectrometer. Carboxylic acids (two compounds on the left) and esters (two com pounds on the right) form fragments mainly due to the cleavage of chemical bonds closest to the C O group. Predict the most prominent peaks in the mass spectra of the two upper compounds. Suggest how you can tell apart these two substances based on their mass spectra 4. Chemical equations of reduction-oxidation (redox) reactions can be quite nontrivial to balance. To do so, you begin with balancing the number of electrons some particles lose in oxidation and other particles gain in reduction. Consider a reaction between potassium permanganate KMnO, and hydrochloric acid HCl), which can be used in a lab to produce chlorine gas Cl2. In this reaction, a MnO; ion is reduced to a Mn2+ ion by getting electrons from Cl and losing its oxygen atoms to bind with H and form water. Calculate the number of electrons in the processes of reduction, MnO; + H+electrons Mn2 H2O (first half-reaction), and oxidation, Cl - electrons Cl (second half-reaction). Then multiply those half-reactions with appropriate numerical factors to equate the number of electrons given and taken in these processes. Then add both half-reactions to get the net equation. Finally, add spectator ions considering that the products include potassium and manganese(II) chlorides. 5. In an aqueous solution of an acid (or a base), there is an equilibrium between the acid HA and the products of its dissociation -H and A: HAH +A. This equilibrium can be described using a dissociation constant: K, Awhere c is molar concen- tration. The values of K. for various acids are known. They show how far the dissociation goes and how strong the acid is. For strong acids, it can be of the order of 10° (molar concentrations are assumed in mol/L, and the units of K, are dropped). In calculations, it is often convenient to use the degree of dissociation, a, which is a fraction of the acid molecules that have dissociated. For acetic acid, K, 1.8-10-3. Evaluate the concentration of hydrogen cations in its 0.010M aqueous solution. Do it rigorously using the degree of dissociation and then approximately assuming that in equilibrium, the concentration of HA)

Answer 1

Question 5.

Given that [HA] = 0.01 M

HA(aq) <----> H⁺(aq) + A^-(aq)

Ka = [H⁺(aq)][A^-(aq)]/[HA(aq)]

Here, Ka = acid dissociation constat = 1.8*10^-5 and [H⁺(aq)] = [A^-(aq)]

i.e. 1.8*10^-5 = [H⁺(aq)]²/0.01

i.e. [H⁺(aq)]² = 1.8*10^-7

i.e. [H⁺(aq)] = 4.243*10^-4 M

Now, pH = -Log[H⁺(aq)] = -Log(4.243*10^-4) = 3.37