Answer:
The volume of 5.0 g CO 2 is 2.6 L CO 2 at STP
Explanation:
STP
STP is currently
0
∘
C
or
273.15 K
, which are equal, though the Kelvin temperature scale is used for gas laws; and pressure is
10
5
.
Pascals (Pa)
, but most people use
100 kPa
, which is equal to
10
5
.
Pa
.
You will use the ideal gas law to answer this question. Its formula is:
P
V
=
n
R
T
,
where
P
is pressure,
V
is volume,
n
is moles,
R
is a gas constant, and
T
is temperature in Kelvins.
Determine moles
You may have noticed that the equation requires moles
(
n
)
, but you have been given the mass of
CO
2
. To determine moles, you multiply the given mass by the inverse of the molar mass of
CO
2
, which is
44.009 g/mol
.
5.0
g CO
2
×
1
mol CO
2
44.009
g CO
2
=
0.1136 mol CO
2
Organize your data
.
Given/Known
P
=
100 kPa
n
=
0.1136 mol
R
=
8.3145 L kPa K
−
1
mol
−
1
https://en.wikipedia.org/wiki/Gas_constant
T
=
273.15 K
Unknown:
V
Solve for volume using the ideal gas law.
Rearrange the formula to isolate
V
. Insert your data into the equation and solve.
V
=
n
R
T
P
V
=
0.1136
mol
×
8.3145
.
L
kPa
K
−
1
mol
−
1
×
273.15
K
100
kPa
=
2.6 L CO
2
rounded to two significant figures due to
5.0 g
Answer link
Doc048
May 18, 2017
I got 2.55 Liters
Explanation:
1 mole of any gas at STP = 22.4 Liters
5
g
C
O
2
(
g
)
=
5
g
44
(
g
mole
)
=
0.114
mole
C
O
2
(
g
)
Volume of 0.114 mole
C
O
2
(
g
)
= (0.114 mole)(22.4 L/mole) = 2.55 Liters
C
O
2
(g) at STP
nuetron, a neutron has no charge.
<span> Hydrofluoric acid is a powerful and dangerous acid. </span>
It would most likely be acid rain
Answer:
37064 J
Explanation:
Data Given:
mass of Steam (m) = 16.4 g
heat released (Q) = ?
Solution:
This question is related to the latent heat of condensation.
Latent heat of condensation is the amount of heat released when water vapors condenses to liquid.
Formula used
Q = m x Lc . . . . . (1)
where
Lc = specific latent heat of condensation
Latent heat of vaporization of water is exactly equal to heat of condensation with - charge
So, Latent heat of vaporization of water have a constant value
Latent heat of vaporization of water = 2260 J/g
So
Latent heat of condensation of water will be = - 2260 J/g
Put values in eq. 1
Q = (16.4 g) x (- 2260 J/g)
Q = - 37064 J
So, 37064 J of heat will be released negative sign indicate release of energy
