2. For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: a. Os(g) + H:O(g) → 2 O:(g) + H2(g) b. 4 NH3(g) + 5 Oz(g) → 4 NO(g) + 6 H;O(g) c. 2 C2H2(g) + 5 0:(g) → 4 CO2(g) + 2 H:O(g) d. CaH;NH2(g) –→ CaHe(g) + NH>(g)
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Number 2 a to d if its okay but if its not okay then, a and b only then Ill ask the other two in separate
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- MATHEMATICAL A catalog in the lab has a recipe for preparing 1 L of a TRIS buffer at 0.0500 M and with pH 8.0: dissolve 2.02 g of TRIS (free base, MW=121.1 g/mol) and 5.25 g of TRIS hydrochloride (the acidic form, MW=157.6 g/mol) in a total volume of 1 L. Verify that this recipe is correct.pH of solution 14.00 12.00 10.00 8.00 6.00 First 4.00 equivalence point 2.00 0 First midpoint Second equivalence point Third midpoint pH = pKa=12.32 Third equivalence point HPO4(aq) + OH(aq) PO4(aq) + H2O(1) Second midpoint pH = pKa = 7.21 H2PO4 (aq) + OH(aq) HPO42 (aq) + H2O(1) Using the Henderson- Hasselback equation, show how to create 2L of a 0.1 M KPhos pH 7.5 buffer using K2HPO4 and KH2PO4. The chart to the left should help you understand what pKa to start with. Show your work. pH = pKa = 2.16 H3PO4(aq) + OH(aq) H2PO4(aq) + H2O(l) 25.0 50.0 75.0 100.0 Volume of NaOH added (mL)SHOW COMPLETE SOLUTION 1. An electrochemical cell consist of 0.2715 Ibs of MnO2, and 0.2524 Ibs of PbO2. A current flow through a 1.1 ohm resistor of electrochemical cell for 45 mins. The reaction takes place at 400k and the total volume of the solution is 504.9 cm^3. Use the standard reduction cell potential table for MnO2 and PbO2: concentration of Mn2+ (2.81 M) and PbO2 (0.95 M) Cell Potential E=0.48 V current when it passes through a 1.1 ohm resistor = 0.44 ampere Determine the following: a. quantity of electricity/charge, Q b. Energy consumed in Joules c. Power in Kilowatts
- Calculation at the pre-equivalence point region of a complexation titration: What is the pSr value of a 50.00 mL solution of 0.01000 M Sr2+ (formation constant, Kf = 5.348 x 108) after the addition of 11.7 of 0.02000 M EDTA which was buffered to pH 11.00? Round your final answer to two places to the right of the decimal point.Result nad Discussion Lead Acetate Reaction: Samples: lysine, cysteine, methionine Reagents: 10% Sodium Hydroxide (NaOH) and Lead Acetate Pb(CH3COO)2 -To 1 ml of the amino acid solution taken in a test tube, add few drops of sodium hydroxide (40%) and boil the contents for 5-10 mins over a bunsen burner. Cool the contents and add few drops of 10% Lead acetate solution and observe.Highlight your values of A,B,C and D. For your question: A mL of B mol/L sodium phosphate solution is combined with C mL of D mol/L calcium bicarbonate. (Calcium bicarbonate is soluble.) A mL B mol/L C mL D mol/L 75.0 ml 0.300 67.5 0.350 Before you begin your reaction, you must accurately produce 1.500 L of your sodium phosphate solution from sodium phosphate trihydrate solid. Write out a procedure to explain all the steps you will take in the lab when making the solution to ensure that your solution concentration is accurate. Please include calculations that show the required mass of solid. Also include the correct names of all equipment used.
- Find AG and AS for the experiment with Urea Standard State/when it melts AG: Mass of Urea (in grams) = 3.64 g Final volume of solution in graduated cylinder = 6.2 mL T= 298.15 K K = [(NH2)2CO]eq R = 0.008314 kJ/mol K AG = -RT In K [(NH2)2CO] = mol of (NH2)2CO/L of solution AS: AG(KJ/mol) = AH(kJ/mol) - TAS(kJ/mol K – times it by 1000) T= 298.15 K AG = ASNaOH.2L X.02mol/L = .004 mol, H3PO4.8L X.15 mol/L = .12 mol H₂PO4 HPO4² + H+ .108 mol start A. If 200 mls of 0.02 M NaoH is added to 800 mls of 0.15 M phosphoric acid buffer at pH 8.2, what is the resultant pH? (pK1 = 2.1, Pk2=7.2, pk3=12.2) add .004 mol OH- end .012mol -.004 .008 mol pH= pka + log [A-]/[HA] = 7.2 + log.112/.008 +.004 .112mol pH=8.35 B. If the same amount of NaOH in part A is added to 800 mls of water, what is the resulting pH? .004mol/1L =.004 [OH-] pOH = 2.4, pH -11.6Benzalkonium chloride (E=0.16) 1.5g qs 100 ml Distilled water 0.9% Nacl solution ad Make isotonic solution Questions: 1. How many ml of distilled water must be added to make the Benzalkonium chloride preparation isotonic? 2. How many ml of 0.9% NaCl must be added to make a 100ml solution?
- i. Write the net ionic equation for this reaction ii. Identify the oxidising and reducing agents from the above reaction 4. (a) Give few comment notes on Different dissociation constants of weak acids observed from the following diagram 14 13- 12 11- K= 1010 K= 10 K= 10 K= 107 K= 10 K= 10 K= 10 K= 10 10 8 pH 6 3 2 1 0.000 0 005 0010 0015 0020 0.025 Vover is given. A +B C+ D 1) Calculate AG for the above react un and indicáte whether the reaction is favorable or unfavorable [A] = 0.9 M 20°C AG° = 4 KJ/mol %3D [B] = 15mM [C] = 4mM [D] = 3 M R= 8.314 J/moleK %3D %3D 4.7 and explain how you know.At Electrochemical Equilibrium, which best describes the distribution of ions in the chamber, on either side of the membrane, when one "mole" of salt (NaCl) is added to one side of a 1 liter chamber, filled with water and divided in half by a membrane permeable only to Na+. 2.0 M CI- and 1.25 M Na+ one side of the membrane; 0.75 M Na+ on the other side of the membrane. 1.0 M CI- and 1.0 M Na+ on both sides of the membrane. 2.0 M CI- and 1.0 M Na+ one side of the membrane; 1.0 M Na+ on the other side of the membrane. 2.0 M CI- and 2.0 M Na+ on one side of the membrane; pure water on the other side of the membrane.