Hi , here's the definition:
Cycloalkanes are cyclic hydrocarbons , meaning that the carbons of the molecule are arranged in the form of a ring.
-Google
Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
Answer:
The gas pressure inside the flask is higher than outside the flask ornthe atmospheric pressure.
Explanation:
Given a U-tube manometer with water in between each arm of U-tube. One end is open to the atmosphere while one end is sealed to a flask filled with a gas. The water in the open end of the tube is experiencing a pressure equal to the atmospheric pressure whereas the water at the end sealed of to the flask is experiencing pressure due to the gas in the flask.
If the water level in both arms of the tube are equal, then the pressure due to the gas is equal to atmospheric pressure.
If the water level in the arm sealed to the gas is higher than that of the open end, The gas pressure is less than the atmospheric pressure.
If the water level in the arm sealed to the flask is lower than the open end, the pressure due to the gas is higher than atmospheric pressure.
From the question, when the water level is higher outside the flask than inside, then the gas pressure inside the flask is higher than the atmospheric pressure.
probably none becuase it going staright unless its going down a hill
Answer:
See explanation
Explanation:
In this case, we have to know where we have to do. We have to remove a "Br" and put a methyl group and for this, we need to use a Gilman reaction. (See figure 1). We have to start with <u>bromomethane</u> that reacts with Li to form <u>methyllithium</u>. Then we have to do the reaction with Copper to obtain the Gilman reagent <u>dimethylcopper</u>. Finally, we have to do the reaction with the halide <u>1-bromocyclohexene</u> to form <u>1-methylcyclohexene</u>.
See figure 1
I hope it helps!