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AlexFokin [52]
3 years ago
5

The solvolysis of t-butyl bromide in methanol yields 2-methylpropene in an E1 reaction (among other products). What is the effec

t on the rate of reaction of tripling the concentration of t-butyl bromide and doubling the concentration of methanol
Chemistry
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

See explanation

Explanation:

For a reaction that proceeds by E1 mechanism, the rate determining step involves the formation of the carbocation.  

The rate of formation of this carbocation depends only on the concentration of the t-butyl bromide since it is the only specie that enters into the rate equation.

Hence, when the concentration of t-butyl bromide is tripled, the rate of reaction is tripled.

Methanol does not enter into the rate equation hence doubling its concentration does not affect the rate of reaction.

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The standard enthapy of combustion of solid urea co(nh2)2 is -632kjmol-1 at 298k​
Setler79 [48]

Answer:

The standard enthalpy of combustion of solid urea ((CO(NH2)2) is -632 kJ mol-1 at 298 K and its standard molar entropy is 104.60 J K-1 mol-1

Explanation:

8 0
3 years ago
72.0 grams of water how many miles of sodium with react with it?
Flura [38]

Answer:

\large \boxed{\text{8.00 mol}}

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ:                  18.02

            2Na + H₂O ⟶ 2NaOH + H₂

m/g:                72.0  

2. Moles of H₂O

\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g  H$_{2}$O}} = \text{3.996 mol H$_{2}$O}

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

\text{Moles of Na} =  \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}

7 0
3 years ago
Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come
Elden [556K]

Answer:

B

Explanation:

For solving this we need a heat balance

Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}

By changing the corresponding relations, we have

m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\

By cancelling similar factor, we obtain

\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\

Which means that the change of temperature in A is twice the change of B

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Answer:

3rd choice

Explanation:

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