If solar illumination contains 800-1000 w/m^2 and a solar cell array has an efficiency of 9.5%, what area of solar cells is needed to obtain enough electrical power for a typical home. a typical home uses 1 kw?

I believe the area of solar cells needed to obtain enough electrical power for a typical home would be (1000 W) / (900 W/m^2) / (0.12) = 9.3 m^2.

gold is my favorite solid.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally