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- Oceanic uptake of carbon dioxide is thus described:CO2 (g) + H2O ⇔ H2CO3, K = [H2CO3]/PCO2 = 3 x 10-2 M atm-1 H2CO3 ⇔ HCO3- + H+, K = [HCO3-][H+]/[H2CO3] = 9 x 10-7 moles/LHCO3- ⇔ CO32 - + H+, K = [CO32 -][H+]/[HCO3-] = 7 x 10-10 moles/LCharge balance equation:[H+] = [OH-] + [HCO3-] + 2[CO32 ] If the CO2 concentration in the atmosphere is 300 ppm, what is the pH of the ocean?Chemistry In an analysis of the content of carbohydrate present in a glycoprotein, the following results were found: 12.6, 11.9, 13.0, 12.7 and 12.5 g of carbohydrate per 100 g of protein. Taking into account that σ is unknown, the confidence interval for the average value at a 90% confidence level of the carbohydrate content is: Select one:to. 12.5 ± 0.4b. 12.5 ± 0.2c. 12.5 ± 0.5d. 12.5 ± 0.3If a bag of fertilizer were labeled as containing 35% K2O, a. What is the analysis when expressed as %K? b. Assume the bag is labeled as 150% P, calculate the percentage P2O5 in the bag.
- Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec The rate law for this reaction is a. rate = k[H2O2][I–][H+] b. rate = k[H2O2][H+] c. rate = k[H2O2]2[I–]2[H+]2 d. None of these e. rate = k[H2O2][I–] f. rate = k[I–][H+]Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec The rate law for this reaction is a. rate = k[H2O2]2[I–]2[H+]2b. rate = k[I–]c. None of thesed. rate = k[I–][H+]e. rate = k[H2O2][I–][H+]f. rate = k[H2O2][H+]Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2][I–][H+]rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec a. The rate law for this reaction is a. rate = k[H2O2][I–][H+] b. rate = k[H2O2]2[I–]2[H+]2 c. rate = k[I–][H+] d. rate = k[H2O2][H+] e. rate = k[H2O2][I–]
- Acetaminophen, a popular drug taken as pain reliever and fever reducer, is produced together with acetic acid from the reaction of 3.05 g 4-aminophenol and 4.1 ml of acetic anhydride. Acetaminophen was extracted at 60% yield. Density of acetic anhydride at 20 C, 1.08 g/ml. Calculate the actual no. of grams of acetaminophen produced. [Determine L.R., E.R.]You have a lot of paperwork laying around. A ream of stacked paper is 500 sheets and is 2.0 inches thick. Let's say you had a mole of paper. (1 mole of paper = 6.022 x 1023 sheets of paper) and it was all stacked up in 10 stacks of equal height, each containing a tenth of a mole of sheets. The ten stacks sit one on top of the other. How high are they? (Hint: find the height of a stack of containing 0.1 mole of paper. USE DIMENSIONAL ANALYSIS!) Here are are some conversions you might need: 2.54 cm = 1 in; around 9.46 x 1015 meters / light yr and 3.26 light yr / parsec.2b) What is the primary consideration?
- Nagyung wanted to determine the heat of dissolution for potassium chlorate (KClO3, MW: 122.55 g/mol) in water. He measured 10 grams of KClO3 and dissolved it in a calorimeter containing 250 g of water. The temperature drop was 3.5 K. What is qH2O? WHat is qrxn? WhaT IS THE change of dissolution of reaction? is the reaction exothermic or endothermic?At 25°C and 1 atm, 0.500 mol/kg solution of NaCl is water has VNaCI = 18.63 cm³/mol and VH2O = 18.062 cm³/mol. Find the volume of a solution prepared by dissolving 0.5000 mol of NaCl in 1000.0 g of water.Chemistry 5 grams of cereal was crushed and dissolved in 25 mL water and placed in a dialysis bag. This sample was dialysed over a week in a beaker which contains 500 mL of water. 5 mL of the dialysate (an aliquote from the 500 mL) was separated to determine the glucose concentration. The glucose concentration was determined as 5 mM. Calculate the amount (% w/w) of glucose present in the cereal. Hint: The molecular weight of glucose is 180.