2.3 dL = .23 liters. d=deci=10^-1 so you move the decimal place back once.
It's D because micro is 10^-6 (move the decimal place back 6 times) 230,000x10^-6=.23 liters or 2.3 dL.
<u>Answer:</u> The mass of solution that the chemistry student should use is 23.4 grams
<u>Explanation:</u>
We are given:
Available mass of isopropenylbenzene = 120. g
Amount of isopropenylbenzene needed by chemistry student = 10.00 g
42.7 % (w/w) solution of isopropenylbenzene.
This means that 42.7 grams of isopropenylbenzene is present in 100 grams of solution.
To calculate the mass of solution for given needed of isopropenylbenzene, we apply unitary method:
For 42.7 grams of isopropenylbenzene, the amount of solution needed is 100 grams
So, for 10.00 grams of isopropenylbenzene, the amount of solution needed will be = 
Hence, the mass of solution that the chemistry student should use is 23.4 grams
Answer:
5.1563867 × 10-9 miles
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Answer:
See answer below
Explanation:
The picture below will show you the final product and mechanism.
In the first step, the NaNH₂ is a strong base, so, this base will substract the hydrogen from carbon 2, to generate a negative charge there, and then, carbon 2 becomes a nucleophyle.
As a nucleophyle it will attack to the CH₃I in the next step, and it will attach to the CH₃.
The second step is just a regular step to reduce the triple bond of the alkyne to alkane or alkene, this will depend on the quantity of the reactant. In this case, an alkene.
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