CHEMISTRY HIGH SCHOOL

Which statements are true of precipitation reactions? Check all that apply.

A. The new compounds are both liquids.
B. Two substances form from one reactant.
C. A solid forms.
D. They are also double-replacement reactions.

Answers

Answer 1
Answer: In chemistry, a precipitation reaction is a double replacement reaction. It involves two reactants and two products. It means that the cations and anions of the compounds interchange with one another to yield two products. But what makes precipitation reactions different is that when two liquid reactants are allowed to react together, a solid substance called precipitate is formed. The solid appears because it is insoluble to the other product in aqueous state.

Therefore, basing on the choices given, precipitation reactions apply to letters C and D.
Answer 2
Answer:

the correct answer to this question would be B and C.


Related Questions

MIDDLE SCHOOL

8. How much enthalpy/heat is transferred when 0.5113 gof ammonia (NH3) reacts with excess oxygen according
| to the following equation:
4NH3 +502 - 4N0+ 6H20
AH = -905.4J​

Answers

When 0.5113 g of ammonia react with excess oxygen, -6.795 J of heat are released.

What is enthalpy?

Enthalpy, a property of a thermodynamic system, is the sum of the system's internal energy and the product of its pressure and volume.

  • Step 1. Write the thermochemical equation.

4 NH₃(g) +5 O₂(g) ⇒ 4 NO(g) + 6 H₂O   ΔH = -905.4J​

  • Step 2. Convert 0.5113 g of NH₃ to moles.

The molar mass of NH₃ is 17.03 g/mol.

0.5113 g × 1 mol/17.03 g = 0.03002 mol

  • Step 3. Calculate the heat released by the reaction of 0.03002 moles of NH₃.

According to the thermochemical equation, -905.4 J are released upon the reaction of 4 moles of ammonia.

0.03002 mol × (-905.4 J/4 mol) = -6.795 J

When 0.5113 g of ammonia react with excess oxygen, -6.795 J of heat are released.

Learn more about enthalpy here: brainly.com/question/11628413

First convert to moles:
0.5113 g / 17 g/mol = 0.0301 mol

Now create a ratio based on the reaction provided to solve for the unknown:

4 NH3 / -905.4 kJ = 0.0301 mol NH3 / x kJ
x = -6.808 kJ
HIGH SCHOOL

25 POINTS In full detail explain why an Iodine atom becomes an Iodide Ion. Be sure to address the following terms in your explanation: valence electrons, octet rule, and noble gases.

Answers

Answer: To achieve noble gas configuration and become stable.

Explanation:

Ions are formed when an atom looses or gains electrons.

If an atom gains electrons, it leads to the formation of negative ions known as anions. If an atom looses electrons, it leads to the formation of positive ions known as cations.

As Iodine has atomic number of 53, it contains 53 electrons which are filled according to Afbau's rule as:

As iodine has 7 valence electrons it is valence shell 5 , it is short of one electron to achieve nearest noble gas configuration of Krypton with atomic number of 54. The elements which follow octet rule are considered to be stable and thus iodine accepts electron to form

EASY AS PIE AND I LIKE PIE

Calcium iodide (CaI2) is an ionic bond, which means that electrons are transferred.  In order for Ca to become the ion Ca2+, the calcium atom must lose 2 electrons. (Electrons have a negative charge, so when an atom loses 2 electrons, its ion becomes more positive.)  In order for I to become the ion I1−, the iodine atom must gain 1 electron. (When an atom gains an electron, its ion will be more negative.)  However, the formula for calcium iodide is CaI2 - there are 2 iodine ions present. This makes sense because the iodine ion has a charge of -1, so two iodine ions have to be present to cancel out the +2 charge of the calcium ion.  Therefore, the calcium atom transfers 2 valence electrons, one to each iodine atom, to form the ionic bond.

IF WRONG, SORRY

COLLEGE

Gold has a molar (atomic) mass of 197 g/mol. consider a 2.47 g sample of pure gold vapor. (a) calculate the number of moles of gold present

Answers

N = given mass/ molar mass.
n = number of moles
given mass = 2.47 g
molar mass = 197 g/mol

n = 2.47 / 197 
n = 0.01253 moles.
I'm sure you wanted to ask more than this. Just put some comments in. I can do the same.
HIGH SCHOOL

What is the term for the dissolving medium in a solution? A.) solvent
B.) solute
C.) solvator
D.) emulsifier

Answers

is the term for dissolving medium in a solution.

Further Explanation:

Solution is a homogeneous mixture of two or more substances. It can exist in any of the phases. Coffee, tea, bleach, lemon juice, and air are some examples of solutions. There are two components in every solution, solute, and solvent. The component that is present in smaller quantity is solute whereas that in larger quantity is known as solvent.

Classification of solutions:

1. Saturated solution

When the maximum amount of solute is dissolved in the solvent, solution becomes saturated solution. Carbonated water saturated with carbon is an example of a saturated solution.

2. Unsaturated solution

When a lesser amount of solute than its maximum capacity has been dissolved in the solvent, the resulting solution is unsaturated solution. Iced tea and coffee are examples of unsaturated solution.

3. Supersaturated solution.

The solution that contains an extra amount of solute than that can be actually dissolved in the solvent is called supersaturated solution.

Since the component present in fewer amounts in the solution is solute while that in larger amounts is solvent. The solution involves the dissolution of solute in the solvent. So solute is the component that gets dissolved and solvent is the component that allows solute to be dissolved in it.

Therefore, the solvent is the dissolving medium in a solution.

Learn more:

  1. The major contribution of Antoine Lavoisier to chemistry: brainly.com/question/2500879
  2. Example of physical change: brainly.com/question/1119909

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Solutions

Keywords: solution, solute, solvent, component, larger quantity, saturated, unsaturated, supersaturated, dissolving medium, homogeneous mixture.

A. Solvent   . . .  . . . . .  . . . . .. . .                                              
Random Questions