ii)State whether you would expect the following Group 13 metal cations.docx

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ii)State whether you would expect the following Group 13 metal cations to act as Lewis acids or bases. Order the cations in terms of softest to hardest. Explain your answer. Al3+, Tl3+, Ga3+, Tl+ (iii) State whether you would expect the following Group 15 hydrides to act as Lewis acids or bases. Order the hydrides in terms of softest to hardest. Explain your answer. PH3, NH3, AsH3 Solution ii) The given group 13 metal cations will act as a Lewis acids , because cations are are electron deficient and act as a electron acceptor. Hard Acids are Lewis acid which are small in size and most likely are the light metal cation with high positive oxidation state like Al 3+ Soft Acids are Lewis acid which are larger in size and most likely are the heavy metal cation associated with low oxidation state like Tl 3+ Tl + Oder of cation from softest to hardest Â Â Tl + < Tl 3+ < Al 3+ (iii) The given Group 15 hydrides will acts as Lewis Base, because Nitrogen, Phosphorus and Arsenic has a lone pair of electron that it can donate. Hard Base are Lewis base anions or neutral molecules which are not easily polarizable like NH 3 Soft Base are Lewis base anions or neutral molecules which are easily polarizable like PH 3 and AsH 3 Order the hydrides in terms of softest to hardest AsH 3 < PH 3 <NH 3 .

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In chapter 2, 5 objectives of memory management were discussed. In Chapter 7 the 5 requirements are discussed. How does each Chapter 2 objective match to a Chapter 7 requirement? Solution After studying chapter 7 learning objectives, you should be able to: Memory management is intended to satisfy the following requirements: Relocation : Programmers typically do not know in advance which other programs will be resident in main memory at the time of execution of their program. Protection: Processes need to acquire permission to reference memory locations for reading or writing purposes. Sharing: Advantageous to allow each process access to the same copy of the program rather than have their own separate copy. Logical organization: Physical organization: Memory available for a program plus its data may be insufficient and overlying allows various modules to be assigned the same region of memory but is time consuming to program. .

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In certain chlorophyll cofactors the Mg2+ is axially coordinated by the S-atom of a methionine. This is odd considering that Mg2+ is considered to be a â€œhard acidâ€ cation. Why would water molecules be a more natural choice for coordination to the Mg2+ than sulfur? What is it then about this protein and its structure that makes the Mg-S coordination happen? Solution Mg 2+ is the bound central metal ion in chlorophyll which is the major photosynthetic pigment.In chlorophyll Mg 2+ is chelated equatorially by tetrapyrrolic macrocycle.One of the two axial position is occupied mainly by histidine residue and other by O-donors,N-donors or S atom. According to Pearson classification Mg 2+belongs to hard ions and it\'s ionic radius is relatively small.It forms stable complexes with O- donors like H2O while less stable complexes with N donors or S atom.Since water is a polar molecule and is electron rich and also the distance between Mg 2+ and water is found to be 2.06 AÂ° which is smaller compared to the others. Also the water and Mg bond is electrostatic in nature.All this makes water and Mg 2+ coordination more stronger and preferred over S atom. S is also being an electron rich ..it can also form bond with Mg 2+. These has been explained on the basis of Density functional theory which gives the distance between Mg 2+ and Water and S atom. .

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In 2012, Shurtleff estimates that warranty costs in the following year will be $25,000. Actual warranty costs in 2013 are only $20,000. What is the effect on the accounting equation when recording estimated warranty costs in 2012? A) Assets decrease B) Stockholdersâ€™ equity increases C) Assets increase D) Stockholdersâ€™ equity decreases E) Liabilities decrease Solution D) Stockholdersâ€™ equity decreases Estimated Warranty cost is recorded as an expense for future warranty work, So it reduces Income, which ultimately decreases Stockholder\'s Equity. .

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In 2000, June, A 75-year-old widow, creates an irrevocable trust naming her five adult grandchildren as the beneficiaries. The assets transferred in trust consist of marketable securities (worth $800,000) and June\'s personal residence (worth $400,000). Bob, June\'s younger brother and a practicing attorney, is designated as the trustee. Other provisions of the trust are as follows: Bob is given the discretion to distribute the income to the beneficiaries based on their need or add it to corpus. He is also given the power to change trust investments and to terminate the trust. The trust is to last for June\'s lifetime or, if sooner, until termination by Bob. Upon termination of the trust, the principal and any accumulated income are to be distributed to the beneficiaries (June\'s grandchildren). For 2000, June files a form 709 to report the transfer in trust and pays a gift tax based on value of $1.2 million ($800,000 + $400,000). After the transfer in trust and up to the time of her death, June continues to occupy the residence. Although she pays no rent, she maintains the property and pays the yearly property taxes. June never discussed the matter of her continued occupancy of the residence with either Bob or the beneficiaries of the trust. Upon June\'s death in 2008, the value of the trust is $2.3 million, broken down as follows: marketable securities and cash ($1.6 million) and residence ($700,000). Shortly thereafter, Bob sells the residence, liquidates the trust, and distributes the proceeds to the beneficiaries. What are the estate taxes consequences of these transaction to June? Solution Since she pays her property taxes, there are no tax consequences on June. However on selling the house, her children will have to pay tax on capital gains from the increase n the value of the house. To avoid this tax, they will have to live in the same house for two years. .

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2. What is a colloid? Why is it difficult to filter a colloid? 3. What is coagulation? What is peptization? 4. Why is the solution of AgNOs added slowly? 5. Why must a large excess of AgNOs be avoided? 6. Why must the silver chloride be protected from the light? 7. For simplicity assume that your unknown chloride is NaCl. Calculate how many mL of the 5% AgNO3 solution must be added to 0.4020 g of the NaCl sample to precipitate all the chloride. Note: 5% means that there are 5 grams of silver nitrate per 100 mL of solution. Solution 2) Colloid : Colloid is a dispersion of small tiny particles in a liquid.. In collods the tiny particles remain suspended in the solvent. rather than settling at the bottom. It is difficult to filter collodal solution because these tiny particles pass through the filter paper. 3) Coagulation : means the aggrigation of oppositely charged particles which settle down under the gravity. Peptization : It is a phenomenon of the formation of a colloidal dispersion from precipitation. 4)Addition of the AgNO3 be done slowly because the addition tends to yield large particles which makes filtration more efficiently .It also helps limit coprecipitation. 5)To avoid coprecipatation the excess of AgNO3 be avoided. 6)Silver chloride must be protected from light as it is unstable. Exposure to light it becomes greyish yellow . Hence, to avoid decompostion, it must be protected from light. .

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2. The Mars Climate Orbiter (formerly the Mars Surveyor 98 Orbiter) was one of two spacecraft in the Mars Surveyor 98 program, the other being the Mars Polar Lander (formerly the Mars Surveyor 98 Lander). The two missions were to study the Martian weather, climate, and water and carbon dioxide badget in order to understand the reservoirs, behavior, and atmospheric role of volatiles and to warch for evidence of long-term and episodic climate changes. The Mars Climate Orbiter was destroyed when a navigation error caused the spacecraft to miss its intended 140-150 km altitude above Mars during orbit insertion, instead entering the Martian atmosphere at about 57 km. The spacecraft was destroyed by atmospherie stresses and friction at this low altitude. A review board found that some data was calculated on the ground in Imperial units (pound-seconds) and eported that way to the navigation team, who were expeeting the data in metric units (newton-seconds). Calculate the conversion between Newton-seconds and pound-seconds. (Hint: pound is the force of 1 slug being accelerated at 1 foot/see2, and 1 slug is 14.59 kg.) s. (Hint: 1 pound is the force of 1 slug being accelerated at 1 foot/see, and 1 slug Solution 1 newtons = 0.224808942443 pounds hence 1 newton -second = Â Â 0.224808942443 pounds -seconds .

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