Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
The top one does because there are more and it’s closer
Answer:
Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements. All elements are pure substances.
Explanation:
PLEASE MARK ME AS BRAINLIEST
Answer: 
kJ
Explanation:
The thermochemical equation for decomposition of ammonium nitrate is:
Given mass= 50.0 kg =
(1kg=1000g)
According to stoichiometry:
1 mole of 
gives = 82.1 kJ of heat
Thus 
of give =
kJ of heat
Thus kJ of heat is evolved from the decomposition of 50.0 kg of ammonium nitrate.
Answer:
Photosynthesis is the answer.
