B, but keep in mind the element sodium is very reactive (though table salts, on the other hand, are not).
<u>Answer:</u> The correct answer is Option 4: U
<u>Explanation:</u>
Activation energy is defined as the energy that is given to a chemical system with potential reactants to produce products. It is represented as 
It is represented as the energy from the reactants to the activated complex.
Reverse reaction is defined as the reaction in which the products formed become reactants and leads to the formation of reactants back.
So, here, the activation energy of the reverse reaction is the energy between the products and the activated complex, which is represented as letter U in the given graph.
Hence, the correct answer is Option 4: U
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
Answer:
1.9 x 10⁻¹³M
Explanation:
Given 0.027M Ca(OH)₂(aq) => 0.027M Ca⁺²(aq) + 2(0.027M) OH⁻(aq)
2(0.027M) OH⁻(aq) = 0.054M OH⁻(aq).
from [H⁺][OH⁻] = 1 x 10⁻¹⁴ at STP Conditions (0°C, 1Atm)
∴{H⁺] = [H₃O⁺] = (1 x 10⁻¹⁴/0.054)M = 1.85 x 10⁻¹³M or 1.9 x 10⁻¹³at 0°C & 1Atm pressure.
Answer:
b. 2 mol of KI in 500. g of water
Explanation:
We have to apply the colligative property of freezing point depression.
The formula is: ΔT = Kf . m . i
As the (Kf . m . i) is higher, then the freezing temperature will be lower.
i refers to the Van't Hoff factor (number of ions dissolved in the solution)
KI → K⁺ + I⁻ (i =2)
Kf is constant so, we have to search for the highest m (molality)
Molality means the moles of solute in 1kg of solvent.
The highest m is option b → 2 mol of KI / 0.5 kg = 4 mol/kg
a. 1 mol of KI / 0.5 kg = 2 mol/kg
c. 1 mol of KI / 1kg = 1 mol/kg
d. 2 mol of KI / 1kg = 2 mol/kg
1000 g = 1kg. In order to determine molality we need to convert the mass (g) of solvent to kg
