compound name compound formula elements present
water H2O hydrogen (H) and oxygen (O)
ammonia NH3 nitrogen (N) and hydrogen (H)
carbon monoxide CO carbon (C) and oxygen (O)
carbon dioxide CO2 carbon (C) and oxygen (O)
<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
<span>Day and night are not exactly of equal length at the time of the March and September equinoxes. The dates on which day and night are each 12 hours occur a few days before and after the equinoxes.</span>
If the formula for density=m/v, you can manipulate this formula to the the mass. After manipulation, you get the equation for mass to be: m=density * volume. With density and volume given, we can find the mass.
mass= (8.9) * 6 = 53.4grams
Answer:
Radius of the interior sphere = 3.847 nm
Explanation:
The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

The first equation becomes

To find ri that satisfies this equation we have to find the roots of the polynomial.
Numerically, it could be calculated that ri=3.847 nm satisfies the equation.
So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.