Bradford Company budgeted 4-350 pounds of material costing $5-7 per po.docx

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Bradford Company budgeted 4,350 pounds of material costing $5.7 per pound to produce 2,175 units. The company actually used 8.000 pounds that cost $5.8 per pound to produce 2,700 units. What is the direct materials quantity variance $14,820 unfavorable $4,995 unfavorable $14,820 favorable $800 unfavorable $435 unfavorable Solution Actual Units ProducedÂ Â 2,700 .

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Brent gets a loan from San Fransisco Bank. Brent signs a standard form note prepared by the bank. The note obligates Brent to pay the bank the amount of the loan, plus interest. The bank is the: A) Maker of the note. B) Drawer of the note. C) Drawee of the note. D) Payee of the note. Solution Promissory note A promissory note is a written document which is prepared at the time of getting loan. It contains terms and conditions of loan including amount of loan, interest rate, repayment schedule , manner of repayment of loan etc. The note is signed by the borrower and then it is kept by lender as a evidence of loan. Promissory note is a negotiable instrument. A note has two original parties. Therefore in the given question San Fransisco is the payee of the notes. Option D .

Brent gets a loan from San Fransisco Bank- Brent signs a standard form.docx

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Calculate the cell potential (Ecell for the following lead concentration cell at 298 K. Pb(+) Pb21 [Pb21 Pb21-0.00700 MlI [Pb21-2.75 M Solution For concentration cell, cathode and anode are same electrode So, Eo = 0 Number of electron being transferred in balanced reaction is 2 So, n = 2 right cell is cathode and left cell is anode use: E = Eo - (2.303*RT/nF) log {[Pb2+] at anode/[Pb2+]at cathode} Here: 2.303*R*T/F = 2.303*8.314*298.15/96500 = 0.0592 So, above expression becomes: E = Eo - (0.0592/n) log {[Pb2+] at anode/[Pb2+]at cathode} E = 0 - (0.0592/2) log (0.007/2.75) E = 7.673*10^-2 V Answer: 7.67*10^-2 V .

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Calculate [OH-] concentration of solutions with the following pH values. Show your work. (a) 10.66 (b) 2.35 (c) 6.02 (d) 13.42 Solution We know that [H + ][OH - ] = 10 -14 Putting log on both sides, - log [H + ] - log [OH - ] = - log (10 -14 ) pH + pOH = 14 pOH = - log [OH - ] [OH - ] = 10 -pOH 1. pOH = 14 - 10.66 = 3.34 [OH - ] = 0.00046 M 2. pOH = 14 - 2.35 = 11.65 [OH - ] = 2.24 x 10 -12 M 3. pOH = 14 - 6.02 = 7.98 [OH - ] = 1.05 x 10 -8 M 4. pOH = 14 - 13.42 = 0.58 [OH - ] = 0.263 M .

Calculate -OH-- concentration of solutions with the following pH value.docx

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Cafeteria plan has following advantages: 1.Employees more likely to appreciate value of their benefit package 2.Flexibility of cafeteria plans help meet various employee needs 3.Cafeteria plans can help control employer costs since employees select only the benefits that they want. Could you give me a example for cafeteria plan\'s advantages? Solution Plan Requirements A cafeteria plan is a written benefit plan maintained by an employer for the benefit of its employees. The plan must allow employees to choose between two or more benefits consisting of cash (or a taxable benefit which is treated as cash) and certain \"qualified benefits.\" See, .

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C++ a) What is a null pointer? b) Give an example of a function that returns a pointer. Solution a) A pointer that is assigned NULL is called a null pointer. The NULL pointer is a constant with a value of zero defined in several standard libraries, including iostream. It is always a good practice to assign the pointer NULL to a pointer variable in case you do not have exact address to be assigned. Example : int *ptr = NULL; // integer pointer pointing to NULL b) // function that returns a pointer to integer variable int *add(int a, int b){ int c = a+ b; int *iptr = &c; // pointer that stores address of c return iptr; } .

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By treating a student as a Solution Invitation of appliction for admission into different streams-->Receipt of Applicaton--> Despatch of admit cards for entrance examination--> Administration of entrance test---> Sortlisting of students based on merit--->Intimation to successful candidates for admission--->Collection of admission fee and admission-->Allocation of Roll numbers and Registration Numbers-->Preparation of Time Table and Teaching--->Conducting Class Test and Semester Examinations--->Evaluation,Compilation and Publication of Results--->Calculation of SGPI &CGPI at the end of each smester--->CGPI at the Final Semester--->Award of Degree and Mark sheet-->Campus Placement based on tests/interviews by visiting companies and organizations--->College Leaving Certificate Issuance--->Registration for Alumni--->Needbased Alumni interaction/Communication. .

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By how much will the pH change if 0.025 mol of HCl is added to 1.00 L of the buffer that contains 0.15 M HC 2 H 3 O 2 and 0.25 M C 2 H 3 O 2 â€“ ? Use K a = 1.8 Ã— 10 â€“5 for HC 2 H 3 O 2 . Solution Moles of HC2H3O2 = 0.15M*1L = 0.15moles Moles of C2H3O2- = 0.25M*1L = 0.25moles Moles of HCl added = 0.025moles HCl will react with C2H3O2- to form HC2H3O2 HCl + C2H3O2- --> HC2H3O2 + Cl- So, moles of C2H3O2- = 0.25moles-0.025moles = 0.225moles Moles of HC2H3O2 = 0.15moles + 0.025moles = 0.175moles pH = pKa + log[C2H3O2-]/[HC2H3O2] Ka = 1.8*10 -5 pKa = -logKa = 4.74 pH = 4.74 + log0.225/0.175 = 4.85 .

By how much will the pH change if 0-025 mol of HCl is added to 1-00 L.docx

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Build a trie of the following terms using: a. A pointer representation (each node points to its children) b. A Leftmost-Child-Right-Sibling representation at, ate, attack, atop, amp, amuse, be, bet, bide Construct an expression tree for the following equation: (a * b) + 9d * -e) Solution Answer Construct an expression tree for the following equation (a*b) +(d*-e) Here expression is the Infix expression , we can create expression tree using operands like +,-,* (a*b) +(d*-e) Infix expression tree + * * a b - e .

Build a trie of the following terms using- a- A pointer representation.docx

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