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1 year ago

Cesium sulfite is formed from the cs cation and the so2−3 anion. What is the correct formula for this compound?

Chemistry

1 answer:

Anna35 [415]1 year ago

3 0

Cesium Sulfite is formed from the Cs cation and the SO₃⁻² anion. The correct formula for this compound is Cs₂SO₃.

<h3>What is Cation ? </h3>

Cations are positively charged ions. It forms when an element lose one or more electrons.

<h3>
What is Anion ? </h3>

Anions are negatively charged ions. It form when an element gains one or more electrons.

Symbol Cs⁺¹ SO₃⁻²

Valency +1 -2

Formula of the compound is Cs₂SO₃.

Thus from the above conclusion we can say that Cesium Sulfite is formed from the Cs cation and the SO₃⁻² anion. The correct formula for this compound is Cs₂SO₃.

Learn more about the Cation and Anions here: brainly.com/question/20005547
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What elements are highlighted:<br> a. Nonmetals<br> b. Metals<br> C. Metalloids

Advocard [28]

Non metals
(It’s both halogens and noble gases ) the other ones which are white(not highlighted) are alkali metals, alkaline earth metals, and transition metals :3

6 0

3 years ago

2p is the correct representation for the sub-shell with n = 2 and l = 1.

boyakko [2]

Explanation:

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

s = 1 orbital

p = 3 orbitals

d = 5 orbitals

f = 7 orbitals

For n = 4

l = 0 to (n-1) = 0 to 3 = (4s , 4p , 4d , 4f)

Number of subshells = 4

Number of orbitals = 1 + 3 + 5 + 7 = 16

The maximum number of electrons the n = 4 shell can contain:

Each orbital can holds upto two electrons, then 16 orbitals will have :

$16\times 2=32$

32 is the maximum number of electrons the n = 4 shell can contain

6 0

3 years ago

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Question 6 of 10

KengaRu [80]

Answer:

option C

Explanation:

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8 0

2 years ago

The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn

Oxana [17]

Answer:

0.193 M

Explanation:

We need to calculate the Zn²⁺ concentration at the cathode where reduction occurs which is the right side in the expression:

Zn2+(aq,0.100 M) ‖ Zn2+(aq,? M) | Zn(s)

Zn²⁺ (aq,?) + 2 e⁻ ⇒ Zn (s)

and oxidation will occur in the anode

Zn (s) ⇒ Zn²⁺ (aq, 0.100 M ) + 2 e⁻

and the overall reaction is

Zn²⁺ (aq,?) ⇒ Zn²⁺ (0.100 M )

The driving force is the difference in concentration and E the electromotive force will be given by

E = Eº - 0.0592/2 log [Zn²⁺ (0.100 M) / Zn²⁺ (M) ]

Plugging the value for E and knowing Eº is cero because we have the same electrodes, we have

17.0 x 10⁻³ = 0 - 0.0592 log 0.100/ X =

- 17.0 x 10⁻³ / 0.0592 = log 0.100 / X

- 0.287 = log (0.100 / X)

Taking inverse log to both sides of the equation

0.516 = 0.100 / X ⇒ X = 0.100 / 0.516 = 0.193 M

4 0

3 years ago

Which of the following would be expected to have the highest viscosity? CH3CH2OH, HOCH2OH, CH3CH2CH3

zaharov [31]

Answer: Out of the given options $HOCH_{2}OH$ is expected to have the highest viscosity.

Explanation:

The resistance occurred in the flow of a liquid substance is called viscosity.

More stronger is the intermolecular forces present in a substance more will be its resistance in its flow. Hence, more will be its viscosity.

For example, $HOCH_{2}OH$ has strong intermolecular hydrogen bonding than the one's present in $CH_{3}CH_{2}OH$ and $CH_{3}CH_{2}CH_{3}$ . This is because two-OH groups are present over here.

Thus, we can conclude that out of the given options $HOCH_{2}OH$ is expected to have the highest viscosity.

5 0

2 years ago