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3 years ago

Describe the two types of vaporization

1 answer:

8 0

Evaporation and transpiration.

The evaporation process uses the sun radiation or sun’s thermal energy for vaporization of a water from the land or soil to be water vapor then in the air. Transpiration on the other hand is the same with evaporation that involves water molecules to be vaporized into the air but it is located in the stoma of the plants. It is like evaporation within plants. This is actually to have homeostasis in plants and still part of the water cycle.

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On the addition of HI

victus00 [196]

Answer:

$CH_3-CH(I)-CH(CH_3)-CH_3$

$I-CH_2-CH_2-CH(CH_3)-CH_3$

Explanation:

Given the compound $CH_2=CH-CH(CH_3)-CH_3$

The steps of reaction:

$HI \longrightarrow H^+ + I^-$

$CH_2=CH-CH(CH_3)-CH_3 + H^+ \longrightarrow CH_3-CH^+-CH(CH_3)-CH_3$

But this intermediate product has a resonance:

$CH_3-CH^+-CH(CH_3)-CH_3 \longleftrightarrow C^+H_2-CH2-CH(CH_3)-CH_3$

The reaction with I-

$CH_3-CH^+-CH(CH_3)-CH_3 + I^- \longrightarrow CH_3-CH(I)-CH(CH_3)-CH_3$

$C^+H_2-CH2-CH(CH_3)-CH_3 + I^- \longrightarrow I-CH_2-CH_2-CH(CH_3)-CH_3$

8 0

3 years ago

How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel

Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

6 0

3 years ago

How many moles of Mg are present in 62.23 grams of Mg? (Show your calculations for full credit.)

mylen [45]

62.23 = 1512.5001499999998 moles

5 0

3 years ago

What would happen to the pressure if the volume were reduced to 0.5 l and the temperature increased to 260 ∘c?

Vsevolod [243]

The ideal gas law. PV=nRT pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273) Original data Pressure 1 atmosphere Volume 1 liter Temp 25C = 298K New data Volume 0.5 liter pressure X Temp 260C = 533K P1v1T1 = P2v2T2 plug and chug. (1)(1)(293) = (x)(0.5)(533) Solve for X, which is the new pressure.

8 0

3 years ago

The air above earth’s surface is at different temperatures. What is responsible for this difference in temperature?

tino4ka555 [31]

The Mesosphere, like the troposphere layer, has a decrease in temperature with altitude because of the decreases in the density of the air molecules. Thermosphere: As the altitude increases, the air temperature increases.

3 0

3 years ago

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