Given:
P1 = 50 kPa
V1 = 5.0 L
P2 = 125 kPa
V2 = ?
Boyle's Law
P1V1 = P2V2
50 * 5 = 125 * V2
250 = 125 * V2
250 / 125 = V2
2 = V2
Answer is letter C. 2.0 L
Answer:
B. It increases.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
It is clear that, at constant P, the volume of the gas (V) is directly proportional to the temperature of the gas.
<em>V ∝ T.</em>
<em></em>
<em>So, If a balloon is heated, the volume of the air in the balloon if the pressure is constant:</em>
B. It increases.
The two types of data are Qualitative and Quantitative data. Qualitative data is the data that cannot be quantified or measured where as quantitative data is quantified that is measured in terms of numbers.
The given statement 17 birds drink from the pond over the course of 2 hours describes the number of birds that drink water over a given amount time. So, it represents quantitative data. Hence the given statement that The following statement describes qualitative data: 17 birds drink from the pond over the course of 2 hours is false.
Answer:
2 moles
Explanation:
Given data:
Moles of water produced =?
Moles of propane = 0.5 mol
Solution:
Chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Now we will compare the moles of propane and C₃H₈.
C₃H₈ : H₂O
1 : 4
0.5 : 4/1×0.5 = 2 mol
Thus 2 mole of water are produced from 0.5 moles of propane.
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: