What is the empirical formula for Hg2(NO3)2

How is oxide different from a neutral oxygen atom

Zepler [3.9K]

Answer: The main difference between oxide and oxygen is that oxide is a chemical compound with at least one oxygen atom while oxygen is an element whose atomic number is 8.

Explanation: let me know if it was right or wrong

4 0

2 years ago

Complete the sentence using ide, ate or Ic. To determine the name of an anion, you take the name of its element and replace the

lara31 [8.8K]

To determine the name of an anion, you take the name of its element and replace the end with "ide".

<h3>What is an anion?</h3>

An anion in chemistry is a negatively charged ion.

Anions are usually formed when a non-metallic atom gains electron(s).

An anion is usually named by taking the elemental name, removing the ending, and adding “ide.

Examples of anions are as follows:

fluoride (F-)
Chloride (Cl-)
Iodide (I-)

Learn more about anions at: brainly.com/question/15578817

#SPJ1

8 0

2 years ago

Given the following thermodynamic data, calculate the lattice energy of LiCl:

tiny-mole [99]

Answer:

$\boxed{\text{-862 kJ/mol}}$

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

(5) Li⁺(g) +e⁻ ⟶ Li(g) 520

(6) Li(g) ⟶ Li(s) -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409

(8) Cl(g) ⟶ ½Cl₂(g) -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻ <u>+349</u>

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862

The lattice energy of LiCl is $\boxed{\textbf{-862 kJ/mol}}$ .

3 0

2 years ago

4. The reaction of silver nitrate and potassium bromide yields silver bromide and potassium nitrate. If

Hatshy [7]

Answer:

1.) AgNO₃

2.) 0.563 moles AgBr

Explanation:

The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).

AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)

Molarity (M) = moles / liters

100 mL = 1 L

AgNO₃

45.0 mL / 100 = 45.0 L

1.25 M = ? moles / 0.450 L

? moles = 0.563 moles

KBr

75.0 mL / 100 = 0.750 L

0.800 M = ? moles / 0.750 L

? moles = 0.600 moles

In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.

4 0

2 years ago

This question deals with waste disposal in the Solutions and Spectroscopy experiment. What should be done to waste solutions con

Korvikt [17]

Answer:

b. It should be dumped in a beaker labeled "waste copper" on one's bench during the experiment.

d. It should be disposed of in the bottle for waste copper ion when work is completed.

Explanation:

Solutions containing copper ion should never be disposed of by dumping them in a sink or in common trash cans, because this will cause pollution in rivers, lakes and seas, being a contaminating agent to both human beings and animals. They should be placed in appropriate compatible containers that can be hermetically sealed. The sealed containers must be labeled with the name and class of hazardous substance they contain and the date they were generated.

It never should be returned to the bottle containing the solution, since it can contaminate the solution of the bottle.

In the Solutions and Spectroscopy experiments there is always wastes.

3 0

3 years ago