Nal is the sodium iodide
The oxidation number of sodium in sodium iodide is 1.<span>Formula: NaIHill system formula: I1Na1CAS registry number: [7681-82-5]Formula weight: 149.894Class: iodideColour: whiteAppearance: crystalline solidMelting point: 660°CBoiling point: 1304°CDensity: 3670 °C</span>
The answer is 9.03 × 10²⁴<span> molecules.
</span><span>Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance.
Make the proportion.
</span><span>6.02 × 10²³ molecules per 1 mol
</span>x per 15 mol
6.02 × 10²³ molecules : 1 mol = x : 15 mol
x = 6.02 × 10²³ molecules * 15 mol * 1 mol
x = 90.3 × 10²³ molecules
x = 9.03 × 10 × 10²³ molecules
x = 9.03 × 10²³⁺¹ molecules
x = 9.03 × 10²⁴ molecules
The overlapping of two s atomic orbitals produces two molecular orbitals, <span>one bonding molecular orbital and one anti-bonding molecular orbital.
Whenever two atomic orbitals overlap according to molecular orbital theory, it will produce one bonding and one anti bonding molecular orbital. Molecular orbital theory is also a way for determining molecular shape wherein electrons are not assigned to character bonds between atoms, however are dealt with as transferring underneath the effect of the nuclei within the entire molecule.</span>
Answer: neutrons have no charge so this might be true
Explanation:
Answer:
I think the density is 64g/cm^3
i did 960 divided by 15 to get 64.