4 (110000)^3 = 4 s^2 0.01s = 10^-5 MoleslitSolution.pdf

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above molecule has 12 signlas. .pdf

Answer(a): We have r(t)=ln19ti+2j+1t2k ------(1) Velocity=v(t)=r\'(t) Now we need todo differentation with respect to( 1) we get r\'(t)=(1/19t)19i+2j+2tk 19 gets cancellled =(1/t)i+2j+2tk Therefore r\'(t)=(1/t)i+2j+2tk----(2) Comparing the above equation with a= xi+yj+zk |a|=sq rt(x2+y2+z2) Therfore x=1/t y=2 z=2t To find ||r\'(t)||=Sq rt[(1/t)2+(2)2+(2t)2 =sq rt[{1/t2+22+(2t)2 =sq rt[1/t2+4+4t2) =sq rt[(1+2t2)2/t2 =eliminating sq root we have =1+2t2/t =1/t+2t Therefore |r\'t|=1/t+2t Answer(b) Therefore speed of t=s(t)= |r\'t| =1/t+2t Answer(C): To find c we need to integrate r\'t and then apply limits Therefore integrating wrt t we have ln(t)i+2tj+t2k Now apply limits upper limit- lower limit to the integral evaluated above ={ln(3)i+2(3)j+(3)2k}-{ln(2)i+2(2)j+(2)2k} =ln3i+6j+9k-ln2i-4j-4k =ln(3/2)i+2j+5k since ln(a)-ln(b)=lna/b Answer:ln(3/2)i+2j+5k Solution Answer(a): We have r(t)=ln19ti+2j+1t2k ------(1) Velocity=v(t)=r\'(t) Now we need todo differentation with respect to( 1) we get r\'(t)=(1/19t)19i+2j+2tk 19 gets cancellled =(1/t)i+2j+2tk Therefore r\'(t)=(1/t)i+2j+2tk----(2) Comparing the above equation with a= xi+yj+zk |a|=sq rt(x2+y2+z2) Therfore x=1/t y=2 z=2t To find ||r\'(t)||=Sq rt[(1/t)2+(2)2+(2t)2 =sq rt[{1/t2+22+(2t)2 =sq rt[1/t2+4+4t2) =sq rt[(1+2t2)2/t2 =eliminating sq root we have =1+2t2/t =1/t+2t Therefore |r\'t|=1/t+2t Answer(b) Therefore speed of t=s(t)= |r\'t| =1/t+2t Answer(C): To find c we need to integrate r\'t and then apply limits Therefore integrating wrt t we have ln(t)i+2tj+t2k Now apply limits upper limit- lower limit to the integral evaluated above ={ln(3)i+2(3)j+(3)2k}-{ln(2)i+2(2)j+(2)2k} =ln3i+6j+9k-ln2i-4j-4k =ln(3/2)i+2j+5k since ln(a)-ln(b)=lna/b Answer:ln(3/2)i+2j+5k.

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1) Vector\'s components in x - direction : 75*cos (35) = 61.436 Vector\'s components in y - direction : 75*sin (35) = 43.0182 2) Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i so magnitude of the horizontal component = 5 3) z = 5 + 12i magnitude of vector = sqrt( 5^2 + 12^2) = 13 direction = atan ( 12/5) =67.3801 degrees resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees. 13 <67.3801 is equivalent to z = 5 + 12i Solution 1) Vector\'s components in x - direction : 75*cos (35) = 61.436 Vector\'s components in y - direction : 75*sin (35) = 43.0182 2) Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i so magnitude of the horizontal component = 5 3) z = 5 + 12i magnitude of vector = sqrt( 5^2 + 12^2) = 13 direction = atan ( 12/5) =67.3801 degrees resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees. 13 <67.3801 is equivalent to z = 5 + 12i.

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annaipowerelectronic

1) Convert angstroms to cm: 4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm 2) Determine the volume of the unit cube: (4.09 x 10-8 cm)^3 = 6.842 10-23 cm3 3) Determine the mass of the metal in the unit cube: 10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g 4) Determine atomic weight (based on 4 atoms per unit cell): 7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to do cross multiplication. x = 108.16 g/mol, so the element is Ag Hope this helps! :-) Solution 1) Convert angstroms to cm: 4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm 2) Determine the volume of the unit cube: (4.09 x 10-8 cm)^3 = 6.842 10-23 cm3 3) Determine the mass of the metal in the unit cube: 10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g 4) Determine atomic weight (based on 4 atoms per unit cell): 7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to do cross multiplication. x = 108.16 g/mol, so the element is Ag Hope this helps! :-).

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annaipowerelectronic

K1+ + CrO42- + Ag1+NO31- Now lets go through the solubility rules and cross out everythingsoluble: K1+ + CrO42- +Ag1+NO31- Group 1A metals are always soluble (Rule 1) Compounds of NO31- are always soluble (Rule2) So we are left with Ag1+ andCrO42- to react with each other asfollows: 2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s) Solution K1+ + CrO42- + Ag1+NO31- Now lets go through the solubility rules and cross out everythingsoluble: K1+ + CrO42- +Ag1+NO31- Group 1A metals are always soluble (Rule 1) Compounds of NO31- are always soluble (Rule2) So we are left with Ag1+ andCrO42- to react with each other asfollows: 2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s).

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Hello. Recently there has been a change in Cramster Rules (I hope you have read the new Terms and Conditions). Cramster now bans users for giving answers directly. We have to send the answers to your inbox after you rate us. That is because people are copying from older threads which have public answers. Hence we can only send the answer after you rate. So please me rate a Lifesaver and I\'ll send the solution to your inbox or email. You\'ve got no other option to get the answer, because no one can give you the answer over here. Even you might be banned for rating a user that gave the answer directly on your question. You need not worry as I have the solution ready in my notebook. Hope you rate me :) Solution Hello. Recently there has been a change in Cramster Rules (I hope you have read the new Terms and Conditions). Cramster now bans users for giving answers directly. We have to send the answers to your inbox after you rate us. That is because people are copying from older threads which have public answers. Hence we can only send the answer after you rate. So please me rate a Lifesaver and I\'ll send the solution to your inbox or email. You\'ve got no other option to get the answer, because no one can give you the answer over here. Even you might be banned for rating a user that gave the answer directly on your question. You need not worry as I have the solution ready in my notebook. Hope you rate me :).

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annaipowerelectronic

first off, you dont\' have hydrogen on the left side of the equation, when something is reduced it GAINS electrons, or gets \" more negative \" in its charge. on the left side, iodine has a charge of + 5 when bonded with oxygen, and -1 alone on the right side, iodine has a charge of 0 zero, oxygen\'s charge remains the same so, Iodine goes from a +5 charge to a zero charge, getting more negative, which is reduced. Solution first off, you dont\' have hydrogen on the left side of the equation, when something is reduced it GAINS electrons, or gets \" more negative \" in its charge. on the left side, iodine has a charge of + 5 when bonded with oxygen, and -1 alone on the right side, iodine has a charge of 0 zero, oxygen\'s charge remains the same so, Iodine goes from a +5 charge to a zero charge, getting more negative, which is reduced..

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4 (110000)^3 = 4 s^2 0.01s = 10^-5 MoleslitSolution.pdf