Answer:
0.13 g
Explanation:
mass of aluminum required = ( Dislocation length) / ( Dislocation density) × (density of metal)
3000 miles to cm ( 1 mile = 160934 cm) = 3000 miles × 160934 cm / 1 mile = 482802000 cm
density of Aluminium = 2.7 g /cm³
dislocation density of aluminum = 10¹⁰ cm³
mass of aluminum required = (482802000 cm × 2.7 g/cm³) / 10¹⁰ cm³ = 0.13 g
Answer:
Ununoctium
Explanation:
It's at the very end of the periodic table
<span>389.88094 amu
First we look up the atomic mass of all elements contained in calcium iodate using the periodic table:
Ca: 40.078
I: 126.90447
O: 15.999
As an intermediate step we calculate the molecular mass of the ion IO3:
126.90447 + 3*15.999 = 174.90147
Then we calculate the mass of one calcium atom and 2 iodate ions:
2*174.90147 + 40.078 = 389.88094 amu</span>
