Answer:
200 mL = 200 cm³
Explanation:
The relationship between cm³ and mL is 1:1.
1 cm³ = 1 mL
Thus, 200 mL is converted to cm³ as follows:
(200 mL)(1 cm³/1 mL) = 200 cm³
The question is incomplete. Complete question is attached below
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Correct Answer:
Option C i.e. I ~ III < IV < V < II
Reason:
During a nucleophilic subsitution reaction of chloroarenes, Cl- group is replaced by an nucleophile like OH-.
Order of reactivity, during such reactions depends on the electron density on carbon atom that is attached to Cl. Lower the electron density, greater will be the reactivity.Among the provided chloroarenes, electron density on C atom will be minimum in case of compound II, because of presence of electron withdrawing group (-NO2) at ortho and para position. Due to this, there will be large number of resonating structures. This signifies greater electron de-localization, and hence largest reactivity for nucleophilic substitution reaction.
Followed by this, compound V will show greater reactivity, due to presence of -NO2 group at para and one of the ortho position. Compound IV will have less number of resonating structures as compared to compound II and V, hence it will display poor reactivity towards nucleophilic substitution reaction.
Finally, compound 1 and III will minimum reactivity towards nucleophilic substitution reaction, because -NO2 group present at meta position (compound III) will not participate in resonance.
Answer:An iodide ion is the ion I−. Compounds with iodine in formal oxidation state −1 are called iodides. This page is for the iodide ion and its salts, not organo iodine compounds. In everyday life, iodide is most commonly encountered as a component of iodized salt, which many governments mandate.
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Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:

The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M