2A 65 wt% Ni-35 wt% Cu alloy is heated to a temperature within the a+ liquid-phaseregion(Figure 2).If the composition of the a phase is 70 wt% Ni, determine(a) the temperature of the alloy(b) the composition of the liquid phase(c) the mass fractions of both phasesComposition (at% Ni)Temperature (°C)о2040609080160010028001500Liquid1453°C26001400Liquidus lineSolidus line24001300+LB1200220011001085C10000204060(Cu)A2000Composition (wt% Ni)80100(Ni)Temperature (°F)

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Chemistry

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Chapter1: Chemical Foundations

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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The thickness of the chromium plate on an auto fender is determined by dissolving a 30.0-cm2section in acid and oxidizing Cr3+ to Cr2O72-with peroxydisulfate. After removing excess peroxydisulfate by boiling, 500.0 mg of Fe(NH4)2(SO4)2•6H2O is added, reducing the Cr2O72-to Cr3+. The excess Fe2+ is back titrated, requiring 18.29 mL of 0.00389 M K2Cr2O7 to reach the end point. Determine the average thickness of the chromium plate given that the density of Cr is 7.20 g/cm3. Note: thickness = volume /area
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The thickness of the chromium plate on an auto fender is determined by dissolving a 30.0-cm2 section in acid and oxidizing Cr3+ to Cr2O7 2-with peroxydisulfate. After removing excess peroxydisulfate by boiling, 500.0 mg of Fe(NH4 )2 (SO4 )2•6H2O is added, reducing the Cr2O7 2- to Cr3+ . The excess Fe2+ is back titrated, requiring 18.29 mL of 0.00389 M K2Cr2O7 to reach the end point. Determine the average thickness of the chromium plate given that the density of Cr is 7.20 g/cm3 . Note: thickness = volume /area
A hypoeutectoid plain-carbon steel that was slow-cooled from the austenitic region to room temperature contains 9.1 wt% eutectoid ferrite. Assume that no change in structure on cooling from just below the eutectoid temperature to room temperature. Calculate the carbon content of the steel. Temperature ( 1000 900 7+ Fe,C 800 a+7 Eutectoid point 0.022 700 600 500- a+FqC 400 0 0.4 100% Fe 0.77 Pearlite Weight percent carbon Fe,C 6.67 1. Let x = the weight percent carbon of the hypoeutectoid steel (shown as a red line). Now, the relation (ratio) between the eutectoid ferrite to the total ferrite and the proeutectoid ferrite is expressed as: [Eutectoid Ferrite] = [Total Ferrite] - [Proeutectoid Ferrite) ... (eq.1) Fill eq.1 by using x. Express with up to 3 significant digits. (NOTE: Please check the rules for significant figures. For example, 0.012 and 1.23 both have 3 significant digits.) [Eutectoid Ferrite]-[1] [Total Ferrite]-[2] [Proeutectoid Ferrite]-[3] *Check the "Question Hint".…
Why is iron generally harder and have higher yield strength (σy) when it is alloyed with carbon? a) Carbon forms an interstitial solid solution in iron and Fe-C bonds are very strong. b) Carbon forms an interstitial solid solution in iron and the C atoms bind dislocations. c) Carbon forms a substitutional solid solution in iron and Fe-C bonds are very strong. d) Carbon forms a substitutional solid solution in iron and the C atoms bind dislocations. e) Carbon is the base element for diamond, which is very hard.

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