Answer:
0.2762M Na+ in the solution
Explanation:
<em>2.07g of sodium iodide Is Dissolved In 50.ML Of A 0.30M...</em>
To solve this question we need to find the moles of sodium iodide, NaI, that are the same than the moles of sodium cation, Na+. The volume in liters of the solution is 0.050L. The molarity is:
<em>Moles NaI = Moles Na+ -Molar mass NaI: 149.89g/mol-</em>
2.07g NaI * (1mol / 149.89g) = 0.01381 moles NaI<em> = Moles Na+</em>
Molarity:
0.01381 moles Na+ / 0.0500L =
0.2762M Na+ in the solution
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Answer:
3.76 g of O₂ are needed to produced 12.5 g of Fe₂O₃
Explanation:
The reaction is: 4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)
4 moles of iron react to 3 moles of oxygen in order to produce 2 moles of iron (III) oxide.
Let's determine the moles of the produced product.
12.5 g . 1mol/ 159.69g = 0.0783 moles
If we assume Iron in excess, we work with the oxygen.
2 moles of Fe₂O₃ are produced by 3 moles of oxygen
Then, 0.0783 moles of Fe₂O₃ might be produced by (0.0783 . 3)/2
0.117 moles.
We convert the moles to mass → 0.117 mol . 32 g/1mol = 3.76 g
Answer:
I believe the answer would be D
Explanation:
The reason why is that I divided 5.75 with 5 which results in 1.15
Answer:
The correct order of increasing reactivity toward nucleophilic acyl substitution is E < D < C < A < F < B.
Explanation:
The stability of the leaving group best determines the manner of reactivity of carboxylates to nucleophilic substitution after the substitution of the nucleophile to the leaving group. The leaving group should, therefore, be protonated with hydrogen ion in the solution to form a stable molecule. From the given list: The leaving group for A, Ethyl thioacetate will be ethanethiol. For B, Acetyl chloride will be Hydrochloric acid. For C, Sodium acetate will be Sodium Hydroxide. For D, Ethyl acetate will be Ethanol. For E, Acetamide will be Ammonia, and for F, Acetic anhydride will be Ethanoic acid. The reactivity of the substitution reaction is dependent on the stability of these leaving groups. The stability of these leaving groups depends on their pKa, and the more the pKa, the lesser the acidity of the leaving group, and the lower the reactivity. Therefore, considering their pKa: A is 8.5, B is -7, C is 13.8, D is 15.9, E is 36, and F is 4.8. When we rearrange this pKa in descending order, we have E, D. C, A, F, B. Which is also the increased reactivity of the nucleophilic acyl substitution.