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3 years ago

What color would a phenolphthalein solution be at ph=11 ?

Chemistry

2 answers:

saveliy_v [14]3 years ago

8 0

Answer:

Pink

Explanation:

The Ph scale is used to determine if a chemical compound is an acid, a base or neutral. The scale ranges from 0 to 14.

Acids turn litmus red and have a ph range between 0 and 6.9. Bases turn litmus blue and have a ph range between 8.2 and 14.

Phenolphthalein is ordinarily colorless however, in a basic solution which gives a ph of 11, it is is gives a pink coloration.

Sloan [31]3 years ago

5 0

Colurs of Phenolphthalein at different pH's are as follow,

When pH is less than zero, means when the conditions are strongly acidic then it imparts Orange Color.

At pH ranging from zero to 8.2 (acidic or weakly acidic conditions) it is colorless.

At pH ranging from 8.2 to 12 (Basic conditions) it imparts Pink color.

At pH greater than 13 (strongly basic) it is again Colorless.

Result:
So, At pH = 11 phenolphthalein solution gives Pink Color.

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“Dry ice” is the solid form of carbon dioxide. Determine the number of mass of CO2 gas in grams that are present in 61.8 L of CO

Fofino [41]

Answer:

Mass = 121 g

Explanation:

Given data:

Mass in gram of CO₂ = ?

Volume = 61.8 L

Pressure = standard = 1 atm

Temperature = 273.15 K

Solution:

Formula:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K × 273.15 k

61.8 L.atm = 22.42 atm.L/ mol × n

n = 61.8 L.atm /22.42 atm.L/ mol

n = 2.76 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 2.76 mol × 44 g/mol

Mass = 121 g

6 0

2 years ago

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

Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the

NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

Answer:

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

Explanation:

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is $ns^{1-2}$

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is $np^{1-6}$

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is $[(n-1)d^{1-10}ns^{0-2}]$

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is $[(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]$ . They are also known as lanthanide and actinide series.

For the given elements:

Option 1: Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^26s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 2: Am

Americium is the 95th element of the periodic table having electronic configuration of $[Rn]5f^{7}6d^07s^2$

As, the last electron is entering the (f subshell), it is a inner transition element.

Option 3: In

Indium is the 49th element of the periodic table having electronic configuration of $[Kr]5s^25p^1$

As, the last electron is entering the p subshell, it is a main group element.

Option 4: Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^56s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 5: As

Arsenic is the 33rd element of the periodic table having electronic configuration of $[Ar]4s^24p^3$

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

Option 6: Se

Selenium is the 34th element of the periodic table having electronic configuration of $[Ar]4s^24p^4$

As, the last electron is entering the p subshell, it is a main group element.

Option 7: Rn

Radon is the 86th element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^6$

As, the last electron is entering the p subshell, it is a main group element.

5 0

3 years ago

Using the periodic table, how would you find elements with chemical properties similar to helium?

rosijanka [135]

The state of the helium in its natural form is gaseous and is a chemical element of colorless aspect and belongs to the group of noble gases. The atomic number of helium is 2. The chemical symbol of helium is He. For the following we focus on those elements and relate it with similar chemical properties. Then we find that; Neon, Hydrogen, Boron and Carbon are related to helium, either by proximity in their atomic number or period or by their group.

3 0

3 years ago

Find the mass of 1.112 moles HF?

Alexeev081 [22]

Answer:

22.25 g

Explanation:

To find the mass, you need to convert moles to grams and get 22.25 g.

7 0

3 years ago