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JulijaS [17]
2 years ago
8

What would be the specific mathematical effect on the reaction rate if you carried out the sodium iodide-in-acetone reactions on

the alkyl halides using an iodide solution half as concentrated? ("Slower" or "faster" is not specific enough.)
Chemistry
1 answer:
Step2247 [10]2 years ago
7 0

Answer:

Slower

Explanation:

The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.

This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.

Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;

Rate = k[alkyl halide] [iodide]

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Answer:

B 1.23 g/cc

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3 years ago
Draw the cis and trans isomers of 2-butene, ch3chchch3. show all hydrogen atoms.
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4 years ago
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3 years ago
The balanced chemical equation below represents the formation of magnesium chloride.
Novay_Z [31]

Answer:

C. 33.6L

Explanation:

Based on the reaction, 2 moles of HCl reacts producing 1 mole of hydrogen.

To solve this question we must find the moles of hydrogen produced using the reaction. Then, with combined gas law (PV = nRT) we can find the volume produced:

<em>Moles H2:</em>

3.00 moles HCl * (1mol H2 / 2mol HCl) = 1.50 moles H2 are produced

<em>Volume:</em>

PV = nRT

V = nRT / P

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = Our incognite</em>

<em>n are moles of the gas = 1.50 moles</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

<em />

V = nRT / P

V = 1.50mol*0.082atmL/molK*273.15K / 1atm

V = 33.6L

<h3>C. 33.6L</h3>
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3 years ago
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Those were the notes my teacher gave me. Hope that helps!! : )
6 0
3 years ago
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