Answer:
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Explanation:
The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:
The answer is (R)-2-chlorobutane.
The reaction take splace through 
mechansim and inversion in configuration happens.
Answer:
The elements in Group 1 (lithium, sodium, potassium, rubidium, cesium, and francium) are called the alkali metals. All of the alkali metals have a single s electron in their outermost principal energy. ... For example, the electron configuration of lithium (Li), the alkali metal of Period 2, is 1s22s1.
It would be at the M phase checkpoint that a cell would be arrested if its chromosome is not properly aligned. It determine if all the chromatids are attached correctly before a cell enters another process. Hope this answers the question.
Answer:
Total Ionic equation:
H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)
Explanation:
Chemical equation:
HNO₃ + NaOH → NaNO₃ + H₂O
Balanced chemical equation:
HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H₂O(l)
Total Ionic equation:
H⁺(aq) + NO₃⁻ (aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + NO₃⁻ (aq)
Net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The NO₃⁻ (aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation
