Problem 2: I have the correct answer I just have to know what formulas he used f

Question

Problem 2: I have the correct answer I just have to know what formulas he used from the formula sheets

An amoeba has 1.00×1016 protons and a net charge of 0.300 pC. If you paired up its protons and electrons, what fraction of the protons would have no electrons?

Circle: [00.500] A=ar2 [00.512] 1 rad = (180 / ) [00.521] 1°-(n / 180) rad [00.531] 1 rev = (2n) rad [00.601] 1 inch = 0.0254 m [00.603] 1 mile = 1609.344 m [00.605] 1 light year = 9.4605284 × 1015 m [00.611] 1 oz. mass0.02835 kg [00.621] 1 year 31558149.76 s [00.623] 1 sidereal day 86164.0196s (00.624 1 sidereal month 2360591.5 s [00.624] 1 synodic month = 25514428 s [00.631] 1 mph 0.44704 m/s [00.641] 1 cal = 4.1868 J [00.643] 1 foot pound = 1.35581791 [00.645] 1 kiloton 4.184 × 1012 J [00.651] 1 horsepower 745.69987 W [00.701] mass of sun = 1.9891 × 1010 kg [00.702] mass of earth = 5.9736 × 1024 kg [00.703] mass of moon 7.349x 10" kg [00.704] mass of electron = 9.10938291 × 10-31 kg [00.706] mass of neutron = 1.674927351 × 10-27 kg [00.501] C 2xr [00.513] 1 rad = (1 / [2 ) rev [00.523] 1°-(1 / 360) rev (00.532] 1 rev 360 [00.602] 1 foot = 0.3048 m [00.604] Bohr radius = 0.52917721 × 10-10 m [00.612] 1 lb. mass0.4536 kg [00.622] 1 day 86400 s [00632] speed of light = 2.99792458 x 108 m/s (exact) (00.642] 1 Cal 4186.8 J [00.644] 1 BTU = 1055.055 9 J [00.705] mass of proton = 1.672621777 × 10-27 kg [00.707] 1 amu [00.712] earth-moon 3.844 × 108 m 1.66053873 x 10" kg [00.711] earth-sun 1.496 × 101 m [00.713] radius of earth = 6.371 x 106 m [00.714] radius of moon = 1.7371 x 106 m [00.801] e = 1.602176565 × 10-19 C (unit of elementary charge) [00.802] ,-8.854187817 × 10-12 F/m [00.803] ,-4m x 10" NA" (exact) [00.803] k = 8.987551798 109 N m2/C2 [00811] h-6.62606957 × 10-34 J s [13.19) k-1.3806488x103J/K [00.812] =1.054571726×10 34Js (13.29] NA602214129x10mol1

Explanation / Answer

Here you have not used any formula from your formula sheet.

Given net charge of protons . To get no of protons you have divided that with proton charge. This has given the information of unpaired protons.

To get fraction you have to divid with total no of protons already given.

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