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navik [9.2K]
3 years ago
11

Which tool was most likely used in a procedure if the lab report shows 300ml of water was used

Chemistry
1 answer:
maxonik [38]3 years ago
3 0
A graduated cylinder, it measures the exact amount of liquid used.
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Which isomer of 1-bromo-2-tert-butylcyclohexane reacts faster when refluxed with potassium tert-butoxide?Draw the structure of t
ivolga24 [154]

Answer:

See explanation and image attached

Explanation:

The reaction of 1-bromo-2-tert-butylcyclohexane with  potassium tert-butoxide is an elimination reaction that occurs by E2 mechanism.

The E2 reaction proceeds faster when the hydrogens are in an antiperiplanar position at an angle of 180 degrees.

This is only attainable in the trans isomer of 1-bromo-2-tert-butylcyclohexane. Hence trans 1-bromo-2-tert-butylcyclohexane reacts faster with potassium tert-butoxide

8 0
3 years ago
Consider the combustion of octane (C8H18)
sesenic [268]
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ --->  16CO₂  + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
 mass of CO₂ formed is 591.8 g
3 0
3 years ago
The worst feeling is when you are hurting so badly inside but not a single tear comes out.
guapka [62]

Answer:

chemistry wow

Explanation:

8 0
3 years ago
Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.
aksik [14]
Molten barium chloride is separeted into two species :
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

<span>The anode is positive and the cathode is negative.</span>

7 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
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