The rusting of iron is an exothermic reaction, 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s). When the coatings are remove to expose metal surface, the system will _. A. Remain constant B. Shift to the left C. Unaffected D. Shift to the right
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The rusting of iron is an exothermic reaction, 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s). When the coatings are remove to expose metal surface, the system will _.
A. Remain constant
B. Shift to the left
C. Unaffected
D. Shift to the rightW
Ps. What happens when the rust is scraped off often?
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- rsity o... ingsbo... kboard... ARED4 ethod? 7... F2 Remaining Time: 1 hour, 22 minutes, 23 seconds. Question Completion Status: A Moving to the next question prevents changes to this answer. Question 1 # The angle of the water molecule (H₂O) is 180 degrees 120 degrees 100 degrees 90 degrees 109 degree A Moving to the next question prevents changes to this answer. MAR 14 80 F3 SA $ 000 F4 tv % F5 NA MacBook Air 22 F6 ∞r F7 45.113The rusting of iron is an exothermic reaction, 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s). When the coatings are remove to expose metal surface, the system will _____. * a. Remain constant b. Shift to the right c. Unaffected d. Shift to the left1. Calculate AH for the reaction: Use the following data: N₂ + O₂2NO 2NO N₂ + 0₂ 2NO+ O₂ → 2NO₂ Unit 7 - Worksheet 4: Hess's Law 2. Calculate AH for the reaction: Use the following data: required! C(graphite) + O₂ → CO₂ C(diamond) + O₂ → CO₂ Use the following data: 3. Calculate AH for the reaction: 2 Al + 1/2O₂ → Al2O3 N₂ + 20₂ >> 2NO₂ 2 Fe + 1½ 0₂ → Fe₂O3 C(graphite) → C (diamond) 2 Al + Fe₂O3 →2Fe + Bat Name: 00 AH = +180 kJ AH=-180kJ 012 Al₂O3 ΔΗ = - 112 kJ AH = - 394 kJ AH = -395 kJ AH = -1669 kJ AH = -824 kJ
- To remove the tarnish (Ag2S) on a silver spoon the following steps are carried out. the spoon is placed in a large pan filled with water so the spoon was totally immersed. A few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolves. Is added. Some aluminum foil is placed at the bottom of the pan in contact with the spoon and then heated the solution to about 80° After a few minutes, the spoon is removed and rinsed with cold water. The tarnish is gone and the spoon regains its original shiny appearance. Describe with equations the electrochemical basis for the procedure. Adding NaCl instead of NaHCO3 would also work because both compounds arestrong electrolytes. What is the added advantage of using NaHCO3? (Hint: Consider the pH of the solution.) What is the purpose of heating the solution? Some commercial tarnish removers containing a fluid (or paste) that is a dilute HCl solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two…The amount of oil and grease from the rivers beside mining sites are quantified by solvent-solvent extraction. The dissolved oil and grease from the organic solvent will then be dried and weighed. What type of gravimetry is described by the statement? O precipitation gravimetry O physical gravimetry thermogravimetry electrogravimetryELECTROCHEMISTRY: Copper was electroplated from a Cu,so, solution onto an electrode with an active surface area of 5.46 cm2. Determine the thickness (mm) of the deposit if the current is kept constant at 4.2 A for 0.8009 hours. (density of Cu = 8.96 g/cm3, MW of Cu = 63.55g/mol). Input values only with 2 decimal places. Do not include the unit.
- Determine the corrosion rate ( mpy ) of steel corresponding to 1µA / cm? of current. Following is the composition of alloys. Cr = 21%, density = 7.1 g/ cm', At.wt. = 52.01 g /mol, number of electron = 1 Ni = 8%, density = 8.9 g/ cm', At.wt. = 58.68 g /mol, number of electron = 2 Fe = 70%, density = 7.86 g/ cm', At.wt. = 55.85 g /mol, number of electron = 2 Note : Faradays constant = 96500 coulombs per mole of electrons. C= constant which includes F and any other conversion factor for unites, for instance, C = 0.129 when corrosion rate is in mpy.How many tabs are to be cut into the filter paper in construction of the apparatus, and how many different metals are to be placed on the apparatus? O There are to be 5 tabs cut, and S metals placed on them. There are to be 6 tabs cut, but only 5 metals placed on 5 of the 6 tabs. There are to be 5 tabs cut, and 7 metals placed on the S tabs, with two metals overlapping O No tabs are to be cut at all, and all 6 metals are to be placed wherever I feel like.Comparing the two quantities based on the given condition, which is greater I or II? or are they equal? Can the quantities be determined given the information? A plant sample was analyzed for Fe I. % w/w Fe II. % w/w FeO
- Balance the following equation by selecting the coefficients in the dropdown boxes.The coefficient for C5H12 isa. 1b. 2c. 3d. 4The coefficient for O2 isa. 1b. 2c. 3d. 4e. 5f. 6g. 7h. 8i. 9j. 10The coefficient for CO2 is a. 1b. 2c. 3d. 4e. 5f. 6g. 7h. 8i. 9j. 10The coefficient for H2O is a. 1b. 2c. 3d. 4e. 5f. 6g. 7h. 8i. 9j. 10The Ksp values of silver chromate Ag2Cro4 and silver iodate Ag(IO3) are given below. Ag2Cro4 Ag(IO3) 1.12 x 1012 3.17 x 10°8 Ksp Based on these Ksp values, which of the following is true? In the solution consisting of 1.00 x10-4 M Ag* and 5.00 x10-5 M Cro42", Ag2Cro4 precipitate will form. In the solution consisting of 1.0 x10-4 MAg* and 1.0 x10-4 M I03", Ag(IO3) precipitate will form. In pure water, the solubility of Ag2CrO4 is lower than the solubility of Ag(IO3). In the solution consisting of 0.200 M CrO42- and 0.200 M 103', Ag2(CrO4) will precipitate first if we add Ag* ions gradually into the above mixture.Thank you for answering^^Determination of Sulfates in water (Analysis) Materials / Equipment needed Standard Methods Interferences (optional)
