Most integrated circuits are made from silicon, which can be obtained from an in
ID: 807296 • Letter: M
Question
Most integrated circuits are made from silicon, which can be obtained from an inexpensive starting material, SiO2. One step in the purification of silicon is to separate it from solid impurities by forming silicon tetrachloride gas. Given the following reactions, what is the overall enthalpy change in converting 1.00 mol of silicon dioxide into pure silicon? Reaction SiO2 (s) 2 C(s) Si impure s) 2. +690 CO(g) si impure s) 2 Cl2(g) SiCl4(g) 657 625 Si(s) a. 1972 kJ +592 b, 1972 kJ e. -625 kJ c. 592 kJ Which of the following shows the reaction associated with the standard heat of formation of a compound? d. F20g) H2(g) 2 HF(g) a. Cao (s) CO2 g) CaCOs(s) e. Cl2(g) 3 F20g) 2 CIFs(g) b. Colg) Cl20g) CoCl2(g) c. 4 C(s, graphite) 4 H20g) C4 Ha(g) In an experiment, 15.0 cm' of glucose (ceH12os, 180.2 gmol, 1.54 gcm undergoes combustion to form carbon dioxide and water. The energy released is completely transferred to grams of water (co 1.00cal og ec), which then experiences a temperature increase of 74.9°C. What is the food value d. 3.73 Cal/g. a. 1.02 Cal/g e. 15.4 Cal/g. b. 20.7 Cal/g c. 5.74 Cal/g. At a temperature, nitrogen gas effuses 4 times faster than which of the following gases? given d. xenon a. fluorine e, chlorine b. bromine C. argon In an experiment, 25.0 mL of a gas with a pressure of 1.00 a is contained in a balloon at 25.00 C. The balloon's is the 0.750 atm at a volume of 31.1 mL. What is the final temperature of the gas under the new conditions? d. 5.02°C a. 23.3°C e. 46.4°C b. 93.4°C -46.5°CExplanation / Answer
1) Add all the reactions: Si(impure) and SiCl4 are present in both product and reactant sides, so can be eliminated.
Thus, SiO2(s) + 2C(s) + 2Cl2(g) + 2Mg(s) ? 2CO(g) + 2MgCl2(s) + Si(s)(pure)
Therefore, detla Ho = -592 kJ
2) c. 4C + 4 H2 ---> C4H8 because the standard heat of formation is defined as the change in energy for one mole of the compound to form from its elements.
3) C6H12O6 + 6O2 -------> 6CO2 + 6H2O
15 cc of glucose = 15*1.54 g and 23.1/180.2 = 0.128 moles
Energy consumed to raise the temperature of 1150g of water by 74.9 C is 1150*1.00*74.9 = 86.135 kCal
That is, 23.1 g of glucose results 86.135 kCal
Therefore, the food value is 3.73 kCal/g. (in question paper the units are printed erroneously)
4) 1.194 = (M/14)1/2
M = 14*1.1942 = 19.96 which is fluorine (given the options)
5) P1V1/T1 = P2V2/T2
T2 = (0.75*31.1)*(298)/(1*25) = 278 K or 5 oC
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