Please show work and everything to recieve A+ rating The equilibrium phase diagr
ID: 1818876 • Letter: P
Question
Please show work and everything to recieve A+ rating
The equilibrium phase diagram for the MgO-FeO system (for problems 10-16, 10-17, 10-24, 10-25, 10-32 and 10-36). The dashed curve represents the solidus for non-equilibrium cooling. Determine the phases present, the compositions of each phase, and the amount of each phase in wt% for the following MgO-FeO ce-ramics at 2000 degree C. (See Figure 10-17.) (i) MgO-25 wt% FeO; (ii) MgO-45 wt% FeO; (iii) MgO-60 wt% FeO; and (iv) MgO-80 wt% FeO. (b) Consider an alloy of 65 wt% Cu and 35 wt% Al. Calculate the composition of the alloy in at%.Explanation / Answer
a) From graph you can see than Mgo-25wt% FeO line and 2000 degree celcius meet below solidus line so phase present will be 'S(Mg,Fe)O' b) From graph you can see than Mgo-45wt% FeO line and 2000 degree celcius meet between liquidus and solidus line so phase present will be 'S(Mg,Fe)O + Liquid'. c) From graph you can see than Mgo-60wt% FeO line and 2000 degree celcius meet at Liquidus line so phase present will be 'mostly liquid' and a little bit of 'S(Mg,Fe)O'. d) From graph you can see than Mgo-80wt% FeO line and 2000 degree celcius meet above liquidus line so phase present will be 'Liquid'. Part 2) The atomic wt. of Aluminium = 26.98 atomic wt. of copper = 63.55 Moles of Al = wt % of al / atomic weight of al = 35 / 26.98 = 1.2973 Moles of Cu = wt % of Cu / atomic weight of Cu = 65 / 63.55 = 1.0228 Atomic % copper = no.moles of copper / (no.moles of copper + no.moles of aluminium) = 1.0228/(1.0228 + 1.2973) = 44.08 % Atomic % Aluminium = 100 - Atomic % copper = 100 - 44.08 ' = 55.92 %