The s orbitals are not symmetrical in shape is a FALSE statement.
An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.
- The atomic orbitals in the atoms of elements differ in shape.
In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.
- All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.
The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.
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brainly.com/question/5087295
Answer:
CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)
Explanation:
According to this question, sodium carbonate reacts with sulfuric acid to form aqueous sodium sulfate, carbon dioxide and water. The balanced chemical equation is as follows:
Na2CO3(aq) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l)
- Next, split compounds that are aqueous into ions.
2Na+(aq) + CO32-(aq) + 2H+(aq) + SO42-(aq) → 2Na+(aq) + SO42-(aq) + CO2(g) + H2O(l)
- Next, we cancel out the spectator ions, which are ions that remain the same in the reactants and products side of a chemical reaction. The spectator ions in this equation are 2Na+(aq) and SO42-(aq).
CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)
- Hence, the balanced ionic equation is as follows:
CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l)
Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
Explanation:
Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).
The equation is show below as:
[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃ ⇒ [(CH₃)₂CH]₂N⁻Li⁺ + CH₃CH₂CH₂CH₃
N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence, LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)
Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)
