1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
sweet [91]
3 years ago
6

Classify each substance as a strong acid, strong base, weak acid, or weak base. drag each item into the appropriate bin.

Chemistry
1 answer:
Zepler [3.9K]3 years ago
6 0
Since you forgot to include the choices for classification, I would just define each of these and tell you the hints that would help you classify them.

Among these acids and bases, its is the strong acids and strong bases that are easily classified. You should note that there are only 7 strong acids existing. All the rest are weak acids. These 7 acids are: HCl, HBr, HI, HClO₃, HClO₄, HNO₃ and H₂SO₄. On the other hand, there are only 8 strong bases; the rest are weak bases. These are the hydroxides of the Group ! and !! metals: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, and Br(OH)₂.

For the weak acids and weak bases, just remember the definitions of Arrhenius, Lewis and Bronsted-Lowry. A weak base are those compounds that accept H⁺ protons, produce OH⁻ ions when solvated and an electron donor. A weak acid are those compounds that donate H⁺ protons, produce H⁺ ions when solvated and an electron acceptor.
You might be interested in
Which of the following is used in pencils?
Mamont248 [21]

Answer:

1st answer:  A. Graphite

2nd answer:  B.  Mercury

Explanation:

3 0
2 years ago
Read 2 more answers
1s^2 2s^2 2p^6 3s^2 3p^6 how many unpaired electrons are in the atom represented by the electron configuration above?
Sedbober [7]
It's a combination of factors:
Less electrons paired in the same orbital
More electrons with parallel spins in separate orbitals
Pertinent valence orbitals NOT close enough in energy for electron pairing to be stabilized enough by large orbital size
DISCLAIMER: Long answer, but it's a complicated issue, so... :)
A lot of people want to say that it's because a "half-filled subshell" increases stability, which is a reason, but not necessarily the only reason. However, for chromium, it's the significant reason.
It's also worth mentioning that these reasons are after-the-fact; chromium doesn't know the reasons we come up with; the reasons just have to be, well, reasonable.
The reasons I can think of are:
Minimization of coulombic repulsion energy
Maximization of exchange energy
Lack of significant reduction of pairing energy overall in comparison to an atom with larger occupied orbitals
COULOMBIC REPULSION ENERGY
Coulombic repulsion energy is the increased energy due to opposite-spin electron pairing, in a context where there are only two electrons of nearly-degenerate energies.
So, for example...
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−− is higher in energy than
↑
↓
−−−−−

↓
↑
−−−−−

↑
↓
−−−−−
To make it easier on us, we can crudely "measure" the repulsion energy with the symbol
Π
c
. We'd just say that for every electron pair in the same orbital, it adds one
Π
c
unit of destabilization.
When you have something like this with parallel electron spins...
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−
It becomes important to incorporate the exchange energy.
EXCHANGE ENERGY
Exchange energy is the reduction in energy due to the number of parallel-spin electron pairs in different orbitals.
It's a quantum mechanical argument where the parallel-spin electrons can exchange with each other due to their indistinguishability (you can't tell for sure if it's electron 1 that's in orbital 1, or electron 2 that's in orbital 1, etc), reducing the energy of the configuration.
For example...
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−− is lower in energy than
↑
↓
−−−−−

↓
↑
−−−−−

↑
↓
−−−−−
To make it easier for us, a crude way to "measure" exchange energy is to say that it's equal to
Π
e
for each pair that can exchange.
So for the first configuration above, it would be stabilized by
Π
e
(
1
↔
2
), but the second configuration would have a
0
Π
e
stabilization (opposite spins; can't exchange).
PAIRING ENERGY
Pairing energy is just the combination of both the repulsion and exchange energy. We call it
Π
, so:
Π
=
Π
c
+
Π
e

Inorganic Chemistry, Miessler et al.
Inorganic Chemistry, Miessler et al.
Basically, the pairing energy is:
higher when repulsion energy is high (i.e. many electrons paired), meaning pairing is unfavorable
lower when exchange energy is high (i.e. many electrons parallel and unpaired), meaning pairing is favorable
So, when it comes to putting it together for chromium... (
4
s
and
3
d
orbitals)
↑
↓
−−−−−
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−
compared to
↑
↓
−−−−−
↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−

↑
↓
−−−−−
is more stable.
For simplicity, if we assume the
4
s
and
3
d
electrons aren't close enough in energy to be considered "nearly-degenerate":
The first configuration has
Π
=
10
Π
e
.
(Exchanges:
1
↔
2
,
1
↔
3
,
1
↔
4
,
1
↔
5
,
2
↔
3
,

2
↔
4
,
2
↔
5
,
3
↔
4
,
3
↔
5
,
4
↔
5
)
The second configuration has
Π
=
Π
c
+
6
Π
e
.
(Exchanges:
1
↔
2
,
1
↔
3
,
1
↔
4
,
2
↔
3
,
2
↔
4
,
3
↔
4
)
Technically, they are about
3.29 eV
apart (Appendix B.9), which means it takes about
3.29 V
to transfer a single electron from the
3
d
up to the
4
s
.
We could also say that since the
3
d
orbitals are lower in energy, transferring one electron to a lower-energy orbital is helpful anyways from a less quantitative perspective.
COMPLICATIONS DUE TO ORBITAL SIZE
Note that for example,
W
has a configuration of
[
X
e
]
5
d
4
6
s
2
, which seems to contradict the reasoning we had for
Cr
, since the pairing occurred in the higher-energy orbital.
But, we should also recognize that
5
d
orbitals are larger than
3
d
orbitals, which means the electron density can be more spread out for
W
than for
Cr
, thus reducing the pairing energy
Π
.
That is,
Π
W
5 0
2 years ago
Read 2 more answers
A sample of the alloy “electrum” is 62% silver and 38% gold by mass; mass of part/
Tanzania [10]

Answer:

c

Explanation:

edge 2022

5 0
2 years ago
What do elements in thr first two columns of the periodic table have in common?
maxonik [38]

The initial two columns of the periodic table make the s-square, and the components in this square share practically speaking that they have a tendency to lose electrons to pick up soundness.

7 0
2 years ago
Read 2 more answers
Platinum crystallizes in a face-centered cubic cell. the density of platinum is 21.4 g/cm3. calculate the radius of the platinum
IgorLugansk [536]
1 mole of platinum has a mass of 195 g therefore 1 atom will have a mass of 195 g /(6.02 ×10^23) = 3.239 × 10^-22 g
Density is given by dividing mass by volume, thus to get volume, mass is divided by density. 
The volume = (3.239 × 10^-22)/21.4 
                    = 1.514 × 10^-23 cm³
But volume of a sphere is given by 4/3πr³
Therefore, r³ = 3.6129 × 10^-24 
                 r   = ∛(3.6129 × 10^-24)
                      = 1.534 × 10^ -8 cm
Therefore, the radius of the platinum atom is 1.534 × 10^-8 cm

7 0
3 years ago
Other questions:
  • Identify, and briefly explain, two key elements needed for fossilization.<br><br>Please Help Me!!!
    6·2 answers
  • What are scientific models used for? Give an example of each of the follwoing types of models: idea, physical, computer
    8·1 answer
  • 4) If a person is riding a bike and want to slow down, they use the brakes. This is an example of which of the following?
    13·1 answer
  • When 13.6 grams of kcl are dissolved in enough water to create a 180-gram solution, what is the solution's concentration, expres
    13·2 answers
  • Y = 4x - 11 and y = -4x + 5*
    14·1 answer
  • Please help ASAP i’ll mark you as brainlister but you have to explain why so I know you not cappin
    8·1 answer
  • Which of the following regions of the periodic table tends to prefer a –1 charge and occupies Group 17?
    8·1 answer
  • If an atom has 10 protons, 10 neutrons, and 10 electrons, what is the atomic mass of the atom?
    12·1 answer
  • What mass is in 5 moles of helium?
    15·1 answer
  • _______ are used because they're economical, easy to process, easy to transport, and the infrastructure exists for their storage
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!