Molar mass H₃PO₄ = 98.0 g/mol
1 mole ----- 98.0 g
? mole ------ 30.0 g
moles = 30.0 * 1 / 98.0
= 0.306 moles
hope this helps!
Answer:
14700J
Explanation:
From the question given, the following were obtained:
M = 100g
ΔT = 35° C
C = 4.2J/g °C
Q=?
The heat transferred can calculated for by using the following equation
Q = MCΔT
Q = 100 x 4.2 x 35
Q= 14700J
Answer:
f the figure is a square with a perimeter of 16 inches, then each side of the square is 4 inchest in length. Area is found by multiplying the length x the width. For a square, the length and width are the same. A = 16 sq
Explanation:
Words that includes names, places, geometric shapes and finally units of measure.
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
