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Levart [38]
3 years ago
15

Enter the symbol of a sodium ion, Na+, followed by the formula of a sulfate ion, SO42−. Separate the ions with a comma only—spac

es are not allowed. Express your answers as ions separated by a comma.
Chemistry
2 answers:
gladu [14]3 years ago
6 0

Na⁺,SO₄²⁻ is the answer

<h3>Further explanation </h3>

An ion is an atom or molecule that has a net electrical charge. There are many ions, one of them are sodium ion and sulfate ion.

SO₄²⁻ or Sulfate is a naturally occurring polyatomic ion that consist of a central sulfur atom surrounded by four oxygen atoms with occured widely in everyday life. Sodium ions are important for regulation of blood and body fluids, transmission of nerve impulses, heart activity, and certain metabolic functions.

Whereas Na⁺ or Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sulfate ion is a very weak base. Therefore sulfate ion undergoes negligible hydrolysis in aqueous solution.

Enter the symbol of a sodium ion, Na⁺, followed by the formula of a sulfate ion, SO₄²⁻. Separate the ions with a comma only—spaces are not allowed. Express your answers as ions separated by a comma. Therefore the answer is: Na⁺,SO₄²⁻

Hope it helps!

<h3>Learn more</h3>
  1. Learn more about sodium ion brainly.com/question/6839866
  2. Learn more about sulfate ion brainly.com/question/2763823
  3. Learn more about ions brainly.com/question/11852357

<h3>Answer details</h3>

Grade:  9

Subject:  Chemistry

Chapter:  Introduction to Mastering Chemistry

Keywords: sodium ion, sulfate ion, ions, Chemistry, symbol

sladkih [1.3K]3 years ago
6 0

The respective symbols of sodium and sulphate ions are \boxed{{\text{N}}{{\text{a}}^ + },{\text{ SO}}_4^{2 - }}.

Further explanation:

The species that are produced after loss or gain of electrons are known as ions. Since ions are charged entities, these can be either positively charged or negatively charged. The positively charged ions are formed by the loss of electrons from the neutral atom and are known as cations. The negatively charged ions are called anions and are formed by the gain of electrons to the neutral atom.

The formation of anion occurs as follows:

 {\text{A}}\left( {{\text{Neutral atom}}} \right) + {e^ - } \to {{\text{A}}^ - }\left( {{\text{Anion}}} \right)

The formation of cation occurs as follows:

 {\text{C}}\left( {{\text{Neutral atom}}} \right) - {e^ - } \to {{\text{C}}^ + }\left( {{\text{Cation}}} \right)

The atomic number of Na is 11 so it has a configuration of 1{s^2}2{s^2}2{p^6}3{s^1}. It easily loses its 3s electron to form {\text{N}}{{\text{a}}^ + } whose configuration becomes 1{s^2}2{s^2}2{p^6}. These ions are beneficial for heart activity, regulation of blood and body fluids and in transmission of nerve impulses. The formation of {\text{N}}{{\text{a}}^ + } occurs as follows:

 {\text{Na}}\left( {1{s^2}2{s^2}2{p^6}3{s^1}} \right) - {e^ - } \to {\text{N}}{{\text{a}}^ + }\left( {1{s^2}2{s^2}2{p^6}} \right)

{\text{SO}}_4^{2 - } is a polyatomic ion that involves a central sulphur atom surrounded by four oxygen atoms and has a charge of -2 on it. These ions are used as cleansing agents in soaps and shampoos.

Learn more:

  1. Balanced chemical equation: brainly.com/question/1405182
  2. Identify the precipitate in the reaction: brainly.com/question/8896163  

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Ionic compounds

Keywords: ions, cations, anions, loss, gain, electrons, Na, Na+, A, C, A-, C+, atomic number, configuration, formation.

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levacccp [35]

Answer:

76.03 °C.

Explanation:

Equation:

C2H5OH(l) --> C2H5OH(g)

ΔHvaporization = ΔH(products) - ΔH (reactants)

= (-235.1 kJ/mol) - (-277.7 kK/mol)

= 42.6 kJ/mol.

ΔSvaporization = ΔS(products) - ΔS(reactants)

= 282.6 J/K.mol - 160.6 J/K.mol

= 122 J/K.mol

= 0.122 kJ/K.mol

Using gibbs free energy equation,

ΔG = ΔH - TΔS

ΔG = 0,

T = ΔH/ΔS

T = 42.6/0.122

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Coverting Kelvin to °C,

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3 years ago
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In reaction a, each sodium atom gives one electron to a chlorine atom in reaction be an isotope of oxygen decays to form an isot
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In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined
faltersainse [42]

<u>Answer:</u>

<u>For A:</u> The standard cell potential of the reaction is 4.4 V

<u>For B:</u> The standard Gibbs free energy of the reaction is -8.50\times 10^5J

<u>For C:</u> The reaction is spontaneous as written.

<u>Explanation:</u>

  • <u>For A:</u>

The given chemical reaction follows:

2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)

The given half reaction follows:

<u>Oxidation half reaction:</u>  Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V ( × 2)

<u>Reduction half reaction:</u>  Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=1.36-(-3.04)=4.4V

Hence, the standard cell potential of the reaction is 4.4 V

  • <u>For B:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

n = number of electrons transferred = 2mol\text{ e}^-

F = Faradays constant = 96500J/V.mol\text{ e}^-

E^o_{cell} = standard cell potential = 4.4 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J

Hence, the standard Gibbs free energy of the reaction is -8.50\times 10^5J

  • <u>For C:</u>

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

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