What term is used to describe the formation of ions in an aqueous solution from a molecular compound

Answers

Answer 1
Answer:

Answer:

Ionization

Explanation:

Molecular compounds are chemical compounds composed of discrete molecules. A molecular compound undergoes ionization when being dissolved in water and the formation of ions are being produced. For example, hydrogen chloride is a molecular compound, when it dissolves in water, ionization is being carried out, and ions are being formed.

\mathbf{HCl \to H^+_((aq)) + Cl^-_((aq))}


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Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? Al2(SO4)3, i = 4 NH4NO3, i = 2 Mg(NO3)2, i = 3 Na2SO4, i = 3 Sucrose, i = 1

Answers

Answer:

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

Explanation:

Hello,

In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that

- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).

- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).

- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).

- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).

Best regards.

Which best illustrates the way in which radiation transfers thermal energy?

#edge2021

Answers

Answer:

it's b.

Explanation:

thank u so much for this. i appreciate it. lol.

First one
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Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them. H2Se

HBr

H2O

HI

Answers

Explanation:

It is known that acidic strength of hydrides of same group tends to increase when we move from top to bottom in a group. On the other hand, acidic strength of hydrides of same period elements increases when we move from left to right in a period.

As both bromine and iodine belongs to the same group. Also, selenium and oxygen are same group elements. Therefore, their acidic strength increases on moving down the group.

Therefore, we can conclude that acidic strength of given compounds from strongest to weakest is as follows.

                HI > HBr > H_(2)Se > H_(2)O

Final answer:

To rank the acids in decreasing acid strength using periodic trends, consider the size, electronegativity, and presence of lone pairs of electrons. HI is the strongest acid, followed by HBr, H2O, and H2Se.

Explanation:

To rank the acids in order of decreasing acid strength using periodic trends, we need to consider the size and electronegativity of the atoms. The larger the atom, the weaker the acid, and the more electronegative the atom, the stronger the acid. Additionally, we can consider the presence of lone pairs of electrons, as they increase the acidity.

  1. HI - Iodine (I) is larger and less electronegative than the other halogens. It also has a lone pair of electrons, making it the strongest acid.
  2. HBr - Bromine (Br) is larger and less electronegative than chlorine (Cl), and it also has a lone pair of electrons. It is the second strongest acid.
  3. H2O - Oxygen (O) is smaller and more electronegative than the halogens. It does not have a lone pair of electrons, making it a weaker acid than the halogens.
  4. H2Se - Selenium (Se) is larger and less electronegative than sulfur (S). However, it does not have a lone pair of electrons, making it the weakest acid.

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A system absorbs 12 J of heat from the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the internal energy ΔEth of the system?

Answers

Answer: The value of change in internal energy of the system is, 40 J.

Explanation : Given,

Heat  absorb from the surroundings = 12 J

Work done on the system = 28 J

First law of thermodynamic : It is a law of conservation of energy in which the total mass and the energy of an isolated system remains constant.

As per first law of thermodynamic,

\Delta U=q+w

where,

\Delta U = internal energy  = ?

q = heat  absorb from the surroundings

w = work done on the system

Now put all the given values in this formula, we get the change in internal energy of the system.

\Delta U=12J+28J

\Delta U=40J

Therefore, the value of change in internal energy of the system is, 40J.

Calculate the molarity of each solution.a. 0.38 mol of lino3 in 6.14 l of solution
b. 72.8 g c2h6o in 2.34 l of solution
c. 12.87 mg ki in 112.4 ml of solution

Answers

Q1)
molarity is defined as the number of moles of solute in 1 L solution 
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L 
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol 
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol 
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL 
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M

The molarities of the given solutions: (a). 0.38 mol of LiNO₃ in 6.14 L of solution has a molarity of 0.062 M. (b). 72.8 g of C₂H₆O in 2.34 L of solution has a molarity of 0.675 M. (c). 12.87 mg of KI in 112.4 mL of solution has a molarity of 0.000688 M.

To calculate the molarity (M) of a solution, you can use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

a. 0.38 moles of LiNO₃ in 6.14 L of solution:

Molarity (M) = 0.38 moles / 6.14 L = 0.062 M

b. 72.8 grams of C₂H₆O (ethyl alcohol) in 2.34 L of solution:

First, you need to convert grams to moles using the molar mass of C₂H₆O.

Molar mass of C₂H₆O = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol

Now, calculate moles of C₂H₆O:

moles = 72.8 g / 46.08 g/mol = 1.58 moles

Molarity (M) = 1.58 moles / 2.34 L = 0.675 M

c. 12.87 mg of KI in 112.4 mL of solution:

First, convert milligrams to grams (1 g = 1000 mg):

12.87 mg = 12.87 g (since 12.87 mg / 1000 = 0.01287 g)

Now, convert mL to liters (1 L = 1000 mL):

112.4 mL = 0.1124 L

Calculate moles of KI:

Molar mass of KI = 39.10 g/mol (for K) + 126.90 g/mol (for I) = 166.00 g/mol

moles = 0.01287 g / 166.00 g/mol = 7.75 × 10⁻⁵ moles

Molarity (M) = (7.75 × 10⁻⁵ moles) / 0.1124 L = 0.000688 M

So, the molarities of the solutions are as follows:

a. 0.062 M

b. 0.675 M

c. 0.000688 M

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At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Answers

We have that from the Question, it can be said that   The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

From the Question we are told

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Generally the equation for constant temperature  is mathematically given as

(C_2)/(C_1)=(P_2)/(P_1)\n\nTherefore\n\nP_2=(P_1C_1)/(C_1)\n\nP_2=(0.22*1.7)/(0.080)\n\nP_2=4.7atm\n\n

Therefore

The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

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Answer: Partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_(He)=K_H* p_(liquid)

where,

K_H = Henry's constant =?

p_(He) = partial pressure = 1.7 atm

Putting values in above equation, we get:

0.080=K_H* 1.7atm\n\nK_H=0.047Matm^(-1)

To find partial pressure of He would give a solubility of 0.730 M

0.730=0.047Matm^(-1)* p_(liquid)

p_(liquid)=15.5atm

Thus partial pressure of He that would give a solubility of 0.730 M is 15.5 atm