Bose Einstein condensation occur at temperature very close to absolute zero i.e 273.15 degrees centigrade.. Under such conditions, large quantity of Boson occupies the lowest quantum state, at which point microscopic quantum phenomena becomes apparent macroscopically. A BEC is formed by cooling a gas of extremely low density about one hundred thousandth the density of normal air, to ultra low temperature.
Answer:
V
=
1.18
L
Explanation:
In order to solve this problem we would use the Ideal Gas Law formula
P
V
=
n
R
T
P
=
Pressure in
a
t
m
V
=
Volume in
L
n
=
moles
R
=
Ideal Gas Law Constant
T
=
Temp in
K
S
T
P
is Standard Temperature and Pressure which has values of
1
a
t
m
and
273
K
2.34
g
C
O
2
must be converted to moles
2.34
g
C
O
2
x
1
m
o
l
44
g
C
O
2
=
0.053
m
o
l
s
P
=
1
a
t
m
V
=
?
?
?
L
n
=
0.053
m
o
l
s
R
=
0.0821
a
t
m
L
m
o
l
K
T
=
273
K
P
V
=
n
R
T
becomes
V
=
n
R
T
P
V
=
0.053
m
o
l
s
(
0.0821
a
t
m
L
m
o
l
K
)
(
273
K
)
1
a
t
m
V
=
1.18
L
Explanation:
Answer:
It's in weight right so the answer should be a scale.
Explanation:
It's a scale because the mesurement is in grams which is measured in weight.
Answer:
Q = 43,000 calories = 180,000 joules = 180 Kilojoules
Explanation:
Heat flow for phase change can be defined as Q = m·ΔHv
ΔHv = Heat of Vaporization for water = 540 calories/gram = 2259 joules/gram = 2.259 Kilojoules/gram
m = mass of object of interest ( in this problem 80 grams steam 100°C converting to 80 grams of water at 100°C
Q = m·ΔHv = 80g x 540 cals/g =43.200 calories x 4.184 joules/cal = 180,748.8 joules (calculator answer) => 180,000 joules (2 sig. figs.) = 180 Kj