Distance=6 x 10^-10\natomic # for iron = 26
131570
This is a simple application of Coulomb's Law.\nThe Electric Field is simply given by the charge on the nucleus (26e) divided by 4*Pi*EpsilonNought*distance^2\nWhere e is 1.6*10^-19 Coulombs
428688
4 Fe(s) + 3 O2(g) ----> 2 Fe2O3(s)\n\nFirst, you have to check if the chemical equation is balanced. In this case, it is. So, what it tells you is that for every 4 mole of iron that is reacted, 1652 kJ will be released to the surrounding. Now, since O2 is in excess, iron will then be your limiting reagent. If you have 3.34 mole of iron, you can use ratio to determine how much energy will be released.\n\nE = (-1652/4) *3.34 = -1379.42 kJ\n\nNote that the "-ve" sign indicates energy is released instead of being absorbed.
550649
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
32900
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
545610
=5.2*5.2*10^-5\n=27.04^10^-5\n=2.704*10^-4
340256
find the distance of the b1 and b2 then plug them in
87941
Average Power = Work/Time\nWork=Force*Distance\nDistance=2*pi*radius=2*pi*(0.31 m)=(1.95 m)\nAverage Power = ((25 N)*(1.95 m))/(1.6 s)=30.4 W
548956
expand it\n\n24=6r-6+4r\n\nbring 6 to the left\n\n30 = 10r\n\ndivide them\n\n3=r
411491
x-6=-25\nx = -25 + 6\nx = -19
410086
total distance = 100 ft (90+10)\nForce lbs=(rho)lb/ft^3*Vol(ft^3)= 37380lbs\n\nWork = Force * distance = 37380*100=3738000 ft*lbs of work
620475
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
442105
First let's find how many moles of each one we have\n\nmole= M*V, so for \nFeCl3 =0.234*25*10^-3= 5.85*10^-3\nNaOH =0.453*42.5*10^-3 =19.25*10^-3\n\nFeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl\n\nso\n\n1mole FeCl3 reacts with 3mole NaOH\n5.85*10^-3 mole react with x\n\nThus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess\nTherefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.\n\nso from the reaction\n\n1 mole FeCl3 gives 1 mole Fe(OH)3\nor better yet, since the molecular weight of Fe(OH)3 is 106.85 \n1 mole FeCl3 gives 106.85 g Fe(OH)3\n5.85*10^-3 mole give x g\nx= 106.85*5.85*10^-3= 0.625 g\n\nWe calculated that 17.55*10^-3 mole NaOH reacted.\nSo we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH\n\nThese are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.\n\nSo the concentration of the excess NaOH is\nC=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M
305475
26 = x + y\ny = 2x +2
55614
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
525538
Intensity varies as the square of the distance.\nThat should help
528149
Correct formula should be as follows:\n\nE = 1/2*I*w^2
176646
By newron's ii law\nMa=F-f\nf=10-10*0.6\nf=4 N\nf= -4 i (vector notation)