Tuesday

October 21, 2014

October 21, 2014

Total # Posts: 26

**Algebra**

Square of the 3rd side is the hypothenuse. If we square root 1025, we'll get 5sqrt(41). 5sqrt(41) - 20^2 = 625 = 25^2
*October 11, 2013*

**Algebra**

The answer is S. A right angle triangle follows Theorem Phytagoras rule. The sum of squares of 2 sides = length of the 3rd side For S, 25^2 + 20^2 = 1025
*October 11, 2013*

**Algebra**

Use distance formula: sqrt[(x1-x2)^2 + (y1-y2)^2] x1,y1 = (-2,2) x2,y2 = (4,-6) Substitute all values inside. The final answer is an even integer. Work it out! :)
*October 11, 2013*

**Chemistry**

The formula is Q = mc(theta), where theta is difference in temperature (final- initial) m = 43g, Cp = 0.90 J/g & theta = 25 degree C Substitute all the values into the equation above: Q = (43)(0.90)(25) = 967.5 J
*October 10, 2013*

**statistics**

Hi Shanice, Numbers in total : 1,2,3,4,5,6,7,8 In the first draw : 2 was selected, therefore the probability of drawing a 2 is 1/8. (since there is only one 2 in the list) In the second draw, 3 was selected. The probability is again 1/8. Therefore the overall probability is: 1...
*October 10, 2013*

**Math**

Hi Helen, this is easy! Solving the literal equation for 'a' means expressing the equation in terms of a. Let us see the solutions step by step: a) Multiplying the denominator a-b with y gives : y(a-b) = xab b)Expanding the parentheses y(a-b) gives ya-yb = xab c)Since ...
*October 10, 2013*

**algebra**

Multiple of 2 = 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30, 32,34,36,38,40,42,44,46,48,50 etc Multiple of 7 = 7,14,21,28,35,42,49,56,63,70 etc So x = 14,28,42,56,70,84,etc You should know the needed x values because you did not mention the range of x
*October 9, 2013*

**Algebra II, i don't get this??**

Hi Mary, Q30 : Yes you are correct. Vertex is the highest point of a graph. Moreover,the f(x) cannot be zero because if f(x) = 0, then 2|x| = -4 |x| = -2 ,which is undefined since |x|is always positive. Q21 : The answer is D. Try to substitute a number of x value and you'...
*October 9, 2013*

**Chemistry**

Hi Dee, this is ez! H3O+ is an hydronium ion, a Lewis acid and OH-, hydroxide ion is a Lewis base. Reaction between hydronium and hydroxide yields water molecules H3O+ + OH- ----> 2H2O This is called a neutralization reaction.
*October 9, 2013*

**math**

Hi Roger, For this problem, you have to cuberoot each of the element (64, x^5 & y^7) separately. For 64 , cuberoot of 64 is 4. x^5 , cuberoot is x^(5/3) y^7, the cuberoot is y^(7/3) Therefore the simplified cube radical form is 4x^(5/3)y^(7/3) Hope this helps you
*October 9, 2013*

**chemistry**

Thank you Dr.Bob, now I know my mistake.
*October 9, 2013*

**chemistry**

Hi Brittney, The solution is as follows: Always start with the basic equation: H2 + O2 -----> H2O(l) Balancing the equation above gives: 2H2 + O2------> 2H2O(l) Since 200 molecules of H2 & O2 are involved,just multiply the equation above with 200, 200(2H2 + O2)---->...
*October 9, 2013*

**algebra**

I disagree that the answer is c < 4. The correct answer is c > 4. Lets see the steps again: x^2 + 4x + c = 0, b^2 - 4ac < 0 4^2 - 4(1)(c) < 0 16 - 4c < 0 16 < 4c 4c > 16 Therefore c > 4
*October 8, 2013*

**Chemistry**

Assume ideal gas law PV = nRT Since the question mentions constant mass and temperature, P1V1 = P2V2 (1.75)(10.7) = P2(20.0L) P2 = ?
*October 8, 2013*

**Chemistry 12**

Use M1V1 = M2V2 12 mols/ L X 1 L/1000 mL X V1 = 0.2 mols/L X 1 L/1000 mL X 500 mL Figure out V1 (in mL) .
*October 8, 2013*

**chemistry**

Easy! Using Q = mc(delta) where delta = difference between final and initial temperature Given m = 47 kg Q = 3.40 X 10^2 kJ Since heat is being absorbed into the system, this is an endothermic reaction. c for water is 4.2 kJ/kg.C Substituting Q,m,c & initial temperature of ...
*September 15, 2013*

**Algebra**

The answer is the same anyway :)
*September 14, 2013*

**Algebra**

Hi Ruth,this is easy! Imagine a right angle triangle. In this case, the length of the hypothenuse is 12'.Given the angle of elevation 68 degrees. Since it is a right angle triangle,one of the remaining two angles is 90 degrees. So the third angle is 180-90-68 = 22 degrees...
*September 14, 2013*

**Math - Triangles**

The answer should be c). The sum of angles for a triangle is 180 degrees. Since this is an isosceles triangle, it must have two equivalent angles. Therefore 180-26 = 154 degrees. When we divide 154 by 2, we get 77 degrees. So the possible degree measure is 78.
*September 10, 2013*

**Chemistry**

Which one of the following cobalt complexes should be the most stable? a. [Co(NH3)6]2+ b. [CO(NH3)6]3+ c. [Co(en)3]3+ d. [CoF6]3– I think the answer is c, but not really sure. Can Dr.Bob explain ? Thanks
*May 20, 2012*

**Chemistry**

If the base sequence along a segment of DNA were TCGTA,what would be the antisense oligonucleotide synthesised from this sequence?
*April 25, 2012*

**chemistry**

For the reaction: H2PO4- + HAsO42- ---> HPO42- + H2AsO4- Which pair exists as base in the equilibrium? H2PO4- & H2AsO4- are proton donors,therefore they are conjugate acids. HPO42- & HAsO42- are proton acceptors,therefore they are the base pairs asked in the question.Is it ...
*March 27, 2012*

**Chemistry**

So,the answers are HAsO42- & HPO4- rite? Because from what I found,HAs04 2- is weakly basic and HPO4 2- is a conjugate base. So the bases are these two species above. M I rite?
*March 25, 2012*

**Chemistry**

The reaction: H2PO4- + HAsO42 ---> HPO4- + H2AsO4- Which one of the sets above lists both of the base species involved in the equilibrium? I'm confused and couldn't identify the conjugate acids and bases properly. Could Dr.Bob explain abt this problem ? Thank you
*March 25, 2012*

**Chemistry**

But why I can't ignore the 2x? Could Dr.Bob please explain?
*March 25, 2012*

**Chemistry**

The following system is at equilibrium,at 699K in a 5L container. H2(g)+I2(g)--->2HI(g) Kc = 54.9 Initially,the system had 2.50 moles of HI.What is the moles of H2 at equilibrium? Can Dr.Bob check if my steps are correct? My solution: Initially,the concentrations of H2 & I2...
*March 25, 2012*

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