1. It will be absorbed by plant. Or remain in soil that .pdf

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mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% = 50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g Solution mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% = 50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g.

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1. It will be absorbed by plant. Or remain in soil that can be absorbed by plant later. Solid
fertilizers is mainly composed of Urea, when it is added with little amount of water to soil, it get
hydrolyzed and converted to ammonia and Co2. This process is normally take 72 hours. Basing
on water content and extra water might cause urea to run off, leading to loss of urea.
CO(NH2)2 + H2O + urease 2NH3 +CO2 (urea)
2. When added more water to the soil, this water will resist the movement of
microorganisms. This is because oxygen is resisted to move freely, due to which microbial
activity is restricted as well. Lack of oxygen will cause to microorganisms metabolism is been
effected. Normal pathways are disrupted due to lack of O2.
Solution
1. It will be absorbed by plant. Or remain in soil that can be absorbed by plant later. Solid
fertilizers is mainly composed of Urea, when it is added with little amount of water to soil, it get
hydrolyzed and converted to ammonia and Co2. This process is normally take 72 hours. Basing
on water content and extra water might cause urea to run off, leading to loss of urea.
CO(NH2)2 + H2O + urease 2NH3 +CO2 (urea)
2. When added more water to the soil, this water will resist the movement of
microorganisms. This is because oxygen is resisted to move freely, due to which microbial
activity is restricted as well. Lack of oxygen will cause to microorganisms metabolism is been
effected. Normal pathways are disrupted due to lack of O2.

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Given: 15.0ml of concentrated (11.6M) HCl into a final volume of 500mL First step you have to convert ml to l (litters) 15.0ml divided by 1000ml=0.015L 500ml divided by 1000ml=0.5L Then, use this formula to solve concentration: M1V1=M2V2 (0.5L)M1=(0.015L)(11.6M) M1=(0.015L)(11.6) divided by (0.5L) M1=0.348M (use 3 sigfigs as a final answer) Final answer is 0.348M Solution Given: 15.0ml of concentrated (11.6M) HCl into a final volume of 500mL First step you have to convert ml to l (litters) 15.0ml divided by 1000ml=0.015L 500ml divided by 1000ml=0.5L Then, use this formula to solve concentration: M1V1=M2V2 (0.5L)M1=(0.015L)(11.6M) M1=(0.015L)(11.6) divided by (0.5L) M1=0.348M (use 3 sigfigs as a final answer) Final answer is 0.348M.

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Given interval is `[0,2pi]` . And given identity is `secx+tanx=1` which we can write as `1/(cosx)+(sinx)/(cosx)=1` òr` , `(1+sinx)/(cosx)=1` or, `(1+sinx)=cosx` or, `1+sinx-cosx=0` or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0` or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0` or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0` or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0` òr,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` . Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our problem. `rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` . So, `x=0` or, `x=2pi` . Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0` `rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)` or, `x/2=(pi/2+pi/4)` or, `x=pi+pi/2` or, `x=3pi/2` . Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval. Solution Given interval is `[0,2pi]` . And given identity is `secx+tanx=1` which we can write as `1/(cosx)+(sinx)/(cosx)=1` òr` , `(1+sinx)/(cosx)=1` or, `(1+sinx)=cosx` or, `1+sinx-cosx=0` or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0` or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0` or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0` or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0` òr,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` . Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our problem. `rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` . So, `x=0` or, `x=2pi` . Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0` `rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)` or, `x/2=(pi/2+pi/4)` or, `x=pi+pi/2` or, `x=3pi/2` . Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..

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Diffrence between totipotent and pluripotent cells are : Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells. Embryonic cells within the first couple of cell divisions after fertilization are the only cells that are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make up the body; embryonic stem cells are considered pluripotent. Solution Diffrence between totipotent and pluripotent cells are : Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells. Embryonic cells within the first couple of cell divisions after fertilization are the only cells that are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make up the body; embryonic stem cells are considered pluripotent..

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Do you want to evaluate 2sin 18 cos 18? There are a lot of apporaches for this problem. Here I show you one approach: Let x = 18 degrees. sin 2x = sin 36 = cos 54 = cos 3x, using co-function property Expand, 2sin x cos x = cos x cos 2x - sin x sin 2x Cancel cos x and let u = sin x = sin 18, 2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x Collect all terms in one side, 4u^2 + 2u - 1 = 0 Apply quadratic formula, u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5)) cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)] 2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further. Solution Do you want to evaluate 2sin 18 cos 18? There are a lot of apporaches for this problem. Here I show you one approach: Let x = 18 degrees. sin 2x = sin 36 = cos 54 = cos 3x, using co-function property Expand, 2sin x cos x = cos x cos 2x - sin x sin 2x Cancel cos x and let u = sin x = sin 18, 2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x Collect all terms in one side, 4u^2 + 2u - 1 = 0 Apply quadratic formula, u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5)) cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)] 2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further..

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Assuming that the system is a closed system, the mass of the system is constant througout the process.Hence the volume of the system doesnot change making it an isochoric process. So, win = 0 kJ/kg Solution Assuming that the system is a closed system, the mass of the system is constant througout the process.Hence the volume of the system doesnot change making it an isochoric process. So, win = 0 kJ/kg.

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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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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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1. 1. It will be absorbed by plant. Or remain in soil that can be absorbed by plant later. Solid fertilizers is mainly composed of Urea, when it is added with little amount of water to soil, it get hydrolyzed and converted to ammonia and Co2. This process is normally take 72 hours. Basing on water content and extra water might cause urea to run off, leading to loss of urea. CO(NH2)2 + H2O + urease 2NH3 +CO2 (urea) 2. When added more water to the soil, this water will resist the movement of microorganisms. This is because oxygen is resisted to move freely, due to which microbial activity is restricted as well. Lack of oxygen will cause to microorganisms metabolism is been effected. Normal pathways are disrupted due to lack of O2. Solution 1. It will be absorbed by plant. Or remain in soil that can be absorbed by plant later. Solid fertilizers is mainly composed of Urea, when it is added with little amount of water to soil, it get hydrolyzed and converted to ammonia and Co2. This process is normally take 72 hours. Basing on water content and extra water might cause urea to run off, leading to loss of urea. CO(NH2)2 + H2O + urease 2NH3 +CO2 (urea) 2. When added more water to the soil, this water will resist the movement of microorganisms. This is because oxygen is resisted to move freely, due to which microbial activity is restricted as well. Lack of oxygen will cause to microorganisms metabolism is been effected. Normal pathways are disrupted due to lack of O2.

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