6.02 kJ>mol. When a small ice cube at -10°C is put into a cup of water at room temperature, which of the following plays a gr
1 answer:
Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice = latent heat of ice
Latent heat of ice = 334 KJ
⇒ = 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.
You might be interested in
Answer:
-) 2-methylbut-2-ene
-) 2-methylbut-1-ene
-) 3-methylbut-1-ene
Explanation:
in this case, the hydration of alkenes is a <u>marknovnikov reaction</u>, this means that the "OH" group would be added in the <u>most substituted carbon</u> of the double bond. (Figure 1)
For 2-methylbut-2-ene the most substituted carbon is the <u>tertiary carbon</u> (the carbon in the right of the double bond), so we will obtain the desired molecule. In 2-methylbut-1-ene the most substituted carbon is again the <u>tertiary carbon</u> (the carbon in the bottom of the double bond), so we will obtain 2-methyl-2-butanol. Finally, for 3-methylbut-1-ene the carbocation would be formed on carbon 3, this is a secondary carbocation. We can obtain a most stable carbocation if we do a <u>hydride shift</u> (Figure 2). With this new molecule is possible to obtain 3-methylbut-1-ene.
