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Levart [38]
2 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]2 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]2 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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The combustion of propane is best described as an
GalinKa [24]
<h3><u>Answer;</u></h3>

Exothermic chemical reaction

<h3><u>Explanation;</u></h3>
  • Exothermic reactions are chemical reactions where the products have less energy that the reactants.
  • Exothermic reactions give off energy, usually in the form of heat, while endothermic reactions absorb energy.
  • The combustion of propane is definitely an exothermic reaction because it generates a lot of heat.

5 0
2 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
2 years ago
Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is
Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

                no. of moles of NaCl = mass of NaCl / molar mass

                no. of moles of NaCl = 0.45g / 58.44 g/mol

               no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0
2 years ago
Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the io
Fudgin [204]

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

8 0
2 years ago
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
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