Answer:
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
Explanation:
NO₃⁻ → NO₂
In left side, Nitrogen acts with +5 by oxidation number
In right side, the oxidation number is +4
This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.
NO₃⁻ + 1e⁻ → NO₂
In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.
NO₃⁻ + 1e⁻ → NO₂ + H₂O
Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
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:
yes
CH3- CH2=CH2 + H2O —> CH3-CH3-CH2-OH
Answer:
III < II < I < IV
Explanation:
Boiling point is determined by the strength of the intermolecular forces in a substance.
n-pentane and 2,2-dimethylpropane are isomers of each other (same molecular formula) and have only single bonds. Both molecules have only London dispersion forces (LDFs), which are the weakest intermolecular forces. The difference in their boiling point is due to the connectivity of the carbon skeleton. n-pentane is a linear hydrocarbon while 2,2-dimethylpropane is a branched hydrocarbon. The linear structure of n-pentane produces a larger surface area between molecules, so the amount of LDFs between the molecules is increased. Thus, n-pentane has a higher boiling point.
Both 1-pentanol and (R)-4-hydroxypentanoic acid have a higher boiling point than the hydrocarbons because they have hydrogen bonding. 1-pentanol has a single alcohol group while (R)-4-hydroxypentanoic acid has both an alcohol group and an acid group, which can also hydrogen bond. Thus, (R)-4-hydroxypentanoic acid has more hydrogen bonding than 1-pentanol and therefore a higher boiling point.
Answer:
2H202(l) --> 2H2O(l) + O2(g)
Explanation:
The balanced equation above represent the decomposition of hydrogen peroxide (H2O2) to yield liquid water (H2O) and liberate oxygen gas (O2) without any heating.
In the laboratory, the rate of decomposition is increased in the presence of manganese (IV) oxide as catalyst.