Answer:
Reason: Gravitational pull of the Moon
Explanation:
Weight of Garfield is 24 lbs on the Earth. He took a flight from the Earth to the Moon where the gravity is 1/6 of the gravity on the Earth.
But when reaches the Moon he found he weighed only 6 lbs. It is due to the gravitational pull of the Moon.
Mass of an object reamins the same at every location while its weight changes due to the change in acceleration due to gravity. It is equal (on Moon) to 1/6 the of acceleration due to gravity on the earth.
Hello. You did not enter answer options. The options are:
a. The OH group was removed from the reactante
b. The OH group was replaced by an H atom.
c. Two atoms were removed from the reactant.
d. An H atom abd a OH group have been removed drom the reactant.
In addition, you forgot to add an image that complements the question. The image is attached below.
Answer:
d. An H atom abd a OH group have been removed drom the reactant
Explanation:
As you can see in the image, the reagent is an alcohol molecule. For the transformation of the alcohol molecule into an alkene molecule, it is necessary that an intramolecular dehydration occurs, that is, it is necessary that a water molecule (two H atoms and an O atom) go out from inside the alcohol and that is exactly what happened, that is, we can say that an H atom and an OH group were removed from the reagent to form the alkene.
Answer:
What will the horizontal axis represent?
Temperature
What will the vertical axis represent?
number of days
What is an appropriate scale?
0-100
Which interval should be used
20
Answer:
The total heat required is 4088.6 J
Explanation:
We have three processes which involve heat absorption. We have to calculate the heat of each process and then to calculate the total heat.
1- liquid ethanol is heated from 25ºC (298 K) to the boiling point 78.5ºc (351.5 K). We use specifi heat of liquid ethanol to calculate the heat absorbed in this part:
H1= m x Sh x ΔT
H1= m x Sh x (Tfinal - Tinitial)
H1= 3.95 g x 2.45 J/g.K x (351.5 K -298 K)
H1= 517.7 J
2- State change: liquid ethanol is vaporized it turns gaseous ethanol The process occur at constant temperature (78.5ºC= 351.5 K). We need the molecular weight of ethanol (2 x 14 + 5 + 16 + 1= 46 g/mol) to cancel mol unit:
H2= m x ΔHvap x 1/Mw
H2= 3.95 g x 40.5 KJ/mol x 1 mol/46 g
H2= 3.477 KJ= 3477 J
3- Gaseous ethanol is heated from 78.5ºC to a final temperature of 95ºC (368 K). We use the specific heat of gaseous ethanol:
H3= m x Sh x ΔT
H3= 3.95 g x 1.43 J/g.K x (368 K - 351.5 K)
H3= 93.2 J
The total heat required is calculated as follows:
Htotal= H1 + H2 + H3
Htotal= 517.7 J + 3477.7 J + 93.2 J
Htotal= 4088.6 J