Answer:
1, just I) color.
Explanation:
Physical properties are the properties that can be observed without changing the composition of a substance, such as color, temperature, density, and boiling point.
A physical change is a change in the substance that only modifies its aggregation state, such as solidification, and boiling.
Chemical properties are the properties that need a reaction to being observed, such as the combustibility, which needs a combustion reaction to being quantified.
When a chemical reaction occurs, and the composition of the substance change, it's a chemical change.
So, the heating copper with carbon is a chemical reaction, and purification by electrolysis is too. Color is the only physical property.
The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
Answer:
By measuring it's Radical velocity using Doppler Phenomenon...
Explanation:
It is done by measuring the absorption spectrum produces by a star as a result of measuring Doppler's relative wavelength. The star with the lower speed must have lower absorption spectrum extent as compared to the faster one.
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
........................................................................................................................
Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
