Answer:
The heat of vaporization relates the amount of heat required to transform a certain phase of a certain amount of substance from liquid to gas. The heat of vaporization of substances can be expressed in terms of joules per gram or joules per mole.
ExplanaDetermine the total amount of heat, q, that is required to evaporate the given amount of water at its boiling point. For this problem, we simply apply the equation,q
=
m
Δ
H
v
a
p
where m is the mass and
Δ
H
v
a
p
is the enthalpy of vaporization of water. We use the following values:
m
=
100
g
Δ
H
v
a
p
=
2.26
k
J
/
g
We proceed with the solution.
q
=
m
Δ
H
v
a
p
=
(
100
g
)
(
2.26
k
J
/
g
)
=
226 k
j tion:
Answer:
1. D. cal......
2.A. iron
3. D
4.2.44j/g°C A
5,Lf=334J/g B
Explanation:
1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal...............
calorie is the unit of energy
#2: A 200 g block of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................
Q=mcdt
1840=0.2*C*(45-25)
C=460J/KgK
if the specific heat capacity is the above then he substance is iron
#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ .................
Q=mcdT
Q=0.1*10*4180
Q=4180j. answer D
.#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................
Q=mcdT
1830=50/1000*C*15
C=2440j/kg/k
change it to j/g°C
2.44j/g°C A
#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g
Q=mLf
Lf=enthalpy of fusion
3340/10=Lf
Lf=334J/g B
Enthalpy of fusion quantity of heat to convert 1 unit mass of a solid to liquid without any noticeable change in temperature.
b. galaxy
c. Kuiper Belt
d. Oort Cloud
Answer:
Foliation is a strong, parallel alignment of coarse mica flakes and/or of different mineral bands in a metamorphic rock.
Explanation:
Foliation is most commonly prevalent in metamorphic rocks. Foliation is the parallel alignment of textural and structural features of a rock.
Differential stress plays a major role on the texture of metamorphic rocks because it forces the mineral constituent of the rocks to align parallel to each other. Foliation can occur in different ways for example a mineral like mica which is usually platy can crystallize in rocks , due to differential stress the mineral grows in such a way it remains parallel to the movement in which the part of the rock slide relative to one another and parallel to the forces applied or the mineral might grow perpendicular to the direction of the compressed stress.
Notice the foliation in the texture of gneiss .You can see the light and dark mineral found in separate or parallel layer.