What are the factors of 210 and 211?
830263
210=1,210,2,105,3,70,42,5,6,35,7,30,14,15,21,10\n211=1,211
885270
Atomic number\n 8\n \nAtomic mass\n 15.999 g.mol -1\n \nElectronegativity according to Pauling\n 3.5\n \nDensity\n 1.429 kg/m3 at 20°C\n \nMelting point\n -219 °C\n \nBoiling point\n -183 °C\n \nVanderwaals radius \n 0.074 nm\n \nIonic radius\n 0.14 nm (-2)\n \nIsotopes\n 4\n \nElectronic shell \n [ He ] 2s 2 2p 4\n \nEnergy of first ionisation\n 1314 kJ.mol -1\n \nEnergy of second ionisation\n 3388 kJ.mol -1\n \nEnergy of third ionisation\n 5300 kJ.mol -1\n \nDiscovered by\n Joseph Priestly in 1774
951454
it is 178 * 177 !\n\nthis can always be computed.
1178018
This may be useful:\n\nAlt + 160 = á\nAlt + 130 = é\nAlt + 161 = í\nAlt + 162 = ó\nAlt + 163 = ú\nAlt + 164 = ñ\nAlt + 165 = Ñ\nAlt + 129 = ü
756921
the answer is\n2Al + N2 ----> 2AlN
503803
3p > 21 + 6\n now adD 21+6 on the right hand side\nthis will give u 27\n3p > 27\nnow this will give u 27 divided by (/) 3\np > 27/3\np> 9\nthe answer is : p>9
794225
210, according to Wilkipedia.
914600
1272 = 2 x 2 x 2 x 3 x 53   ...prime factors\n2, 3, 4, 6, 8, 12, 24, 53, 106, 159, 212, 318, 424, 636,   ...all factors\noh, I forgot, and 1 and 1272.
104563
My answer is:\n\n.217N
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
352698
7 & 103\n\nYou can go here to find the prime factors for any number:\nhttp://www.acme.com/software/factor/
7508
No, that cannot be an answer because 13*2=26 which is not included in the result.\nI think it has to do something with squares
669514
no, it takes 10 decimeters to =1meter so 21 meters is 210 decimeters
776529
The number of protons in the fission product must be 92-55 = 37.  The number of neutrons must be 1+235-143-3 = 90.  Look up in the periodic table to see what element 37 is. It is virtually certain that 90 neutrons are too many for stability, and the product will be a beta emitter.
1075835
20.\n\n40 - 20 = 20.\n\nThe mass number is the number of protons and neutrons combined. The atomic number is the number of protons.
907838
equations a and b are the length and width.\n200 =2a+pi*b\nA(max)=a*b\nAmax=(200a-2a^2)/pi\ndifferentiate set to zero to find max\n\na=50, b=31.82
350595
C whether this helps You,\nhttp://web.umr.edu/~kpeaslee/classes/Met221/KeyHW2.pdf#search='measuring%20SDAS%20in%20Aluminum%20alloy%20casting'\n\nhttp://www.aluplanet.com/eng/info_eco_doc_PO.asp?Mnu=141&Itm=163&Grp=0&Tgl=162&Doc=3804