B. the shape of a molecule determines its properties and interactions
17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).
![17.8 g \times \frac{1mol}{122.44g} = 0.145 mol](https://tex.z-dn.net/?f=17.8%20g%20%5Ctimes%20%5Cfrac%7B1mol%7D%7B122.44g%7D%20%3D%200.145%20mol)
Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.
![0.145 mol \times \frac{6.02 \times 10^{23}molecules }{mol} = 8.73 \times 10^{22}molecules](https://tex.z-dn.net/?f=0.145%20mol%20%5Ctimes%20%5Cfrac%7B6.02%20%5Ctimes%2010%5E%7B23%7Dmolecules%20%20%7D%7Bmol%7D%20%3D%208.73%20%5Ctimes%2010%5E%7B22%7Dmolecules)
NaClO₄ is a strong electrolyte that dissociates according to the following equation.
NaClO₄ ⇒ Na⁺ + ClO₄⁻
The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:
![8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion Na^{+} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion Na^{+}](https://tex.z-dn.net/?f=8.73%20%5Ctimes%2010%5E%7B22%7DmoleculeNaClO_4%20%5Ctimes%20%5Cfrac%7B1%20ion%20Na%5E%7B%2B%7D%20%7D%7B1moleculeNaClO_4%7D%20%3D%208.73%20%5Ctimes%2010%5E%7B22%7Dion%20Na%5E%7B%2B%7D)
The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:
![8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion ClO_4^{-} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion ClO_4^{-}](https://tex.z-dn.net/?f=8.73%20%5Ctimes%2010%5E%7B22%7DmoleculeNaClO_4%20%5Ctimes%20%5Cfrac%7B1%20ion%20ClO_4%5E%7B-%7D%20%7D%7B1moleculeNaClO_4%7D%20%3D%208.73%20%5Ctimes%2010%5E%7B22%7Dion%20ClO_4%5E%7B-%7D)
The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:
![8.73 \times 10^{22}ion ClO_4^{-} \times \frac{1 atomCl }{1ion ClO_4^{-}} = 8.73 \times 10^{22}atom Cl](https://tex.z-dn.net/?f=8.73%20%5Ctimes%2010%5E%7B22%7Dion%20ClO_4%5E%7B-%7D%20%5Ctimes%20%5Cfrac%7B1%20atomCl%20%7D%7B1ion%20ClO_4%5E%7B-%7D%7D%20%3D%208.73%20%5Ctimes%2010%5E%7B22%7Datom%20Cl)
The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:
![8.73 \times 10^{22}ion ClO_4^{-} \times \frac{4 atomO }{1ion ClO_4^{-}} = 3.49 \times 10^{23}atom O](https://tex.z-dn.net/?f=8.73%20%5Ctimes%2010%5E%7B22%7Dion%20ClO_4%5E%7B-%7D%20%5Ctimes%20%5Cfrac%7B4%20atomO%20%7D%7B1ion%20ClO_4%5E%7B-%7D%7D%20%3D%203.49%20%5Ctimes%2010%5E%7B23%7Datom%20O)
17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
You can learn more Avogadro's number here: brainly.com/question/13302703
Periodic table<span> of the </span>elements, in chemistry, the organized<span> array of all the chemical </span>elements<span> in order of increasing atomic number—i.e., the total number of protons in the atomic nucleus.</span>
The frequency of this line of vanadium is 9.38 x10 ^14 Hz.
Emission spectrum shows how the electron of an atom goes or moves from a higher to a lower energy level.
Now The energy of a photon is given by
E = hc/λ
where
h = Plank's constant = 6.626 x 10⁻³⁴ J s
c = speed of light= 3 x 10⁸ m/s
λ = wavelength = 318.5 x 10⁻⁹ m
Solving
E = (6.626 x 10⁻³⁴ J s x 3 x 10⁸ m/s) / 318.5 x 10⁻⁹ m
E =6.2166 x10 ^-19 J
Also, we know that energy is related to frequency by the equation
E =hf
Where;
h = Planks's constant
f = frequency of photon
Making frequency subject of the formulae
f = E/h
f =6.2166 x10 ^-19 J/ 6.626 x 10⁻³⁴ J s
f = 9.38 x10 ^14 Hz
See similar question and solution here:brainly.com/question/24630281
The earths rotation. The earth rotates and as it rotates, the sun moves across the sky