Answer:
The unit cell edge lenght in pm is equal to 361 pm
Explanation:
Data provided:
ρ=Copper density=8.96 g/cm3
Atomic mass of copper=63.54 g/mol
Atoms/cell=4 atoms (in theory)
Avogadro's number=6.02x
atoms/mol
Since copper has a cubic structure, its cell volume is equal to 
, which can be obtained through the relationship:
cell volume=
Substituting the values:
cell volume=
clearing, we have:
a=![\sqrt[3]{4.71x10^{-23}cm^{3} }=3.61x10^{-8}cm](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4.71x10%5E%7B-23%7Dcm%5E%7B3%7D%20%20%7D%3D3.61x10%5E%7B-8%7Dcm)
We convert from centimeter to picometer, 1cm=1x
pm
a=
Copper ions, nitrate ions, and water is in the beaker
Traditionally they include boron from group 3A, silicon and germanium in group 4A, aresnic and antimony in group 5A and tellurium from group 6A, although sometimes selenium, astatine, polonium and even bismuth have also been considered as metalloids. Typically metalloids are brittle and show a semi-metallic luster.
The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. Five elements are less frequently so classified: carbon, aluminium, selenium, polonium, and astatine.
I think the answer is tenfold
hope this helps :)
F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
