What is the molarity of a solution in which 250 grams of NaCl are dissolved in 2.9 L of water?
1 answer:
Answer:
About 1.48 M.
Explanation:
The formula for molarity is mol/L.
So firstly, you must find the amount of moles in 250 grams of NaCl.
I do this by using stoichiometry. First, I find how nany grams are in a single mole of NaCl. This is around 58.44 grams/mole. Now that I know this, I can now use a stoich table. (250 g NaCl * 1 mol NaCl / 58.44 g NaCl). I plug this into my calculator.
I get that 250 grams of NaCl is equal to about 4.28 moles.
Now I just plug into the formula!
4.28 moles/2.9 L = about 1.48
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Answer:
IV
Explanation:
The complete question is shown in the image attached.
Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.
Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.
Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.
