Silver oxide a compound, an element, homogeneous mixture or heterogeneous mixture

Answers

Answer 1

Answer:

Element

Explanation:

Answer 2

Answer:

Silver oxide is a chemical compound

Explanation:

Silver oxide is the chemical compound with the formula Ag2O.


Related Questions

What's the pollen tube also known as

Answers

Answer:

The stigma.

Explanation:

You're welcome. :)

The mass spectrum of compound A shows the molecular ion at m/z 85, an M +1 peak at m/z 86 of approximately 6% abundance relative to M, and an M+2 peak at m/z 87 of less than 0.1% abundance relative to M. Assuming that compound A has only C, H, and one N atoms, determine the molecular formula, and then draw a possible structure if compound A has IR absorption at 1620-1680 cm^-1 but not at 3010-3090 cm^-1 .

Answers

Answer:

(1). CH3-CH=N-CH2-CH2-CH3.

(2). CH3-CH2-CH =N-CH2-CH3.

(3). CH3-CH2-CH2-CH=N-CH3.

(3). CH3-CH2-CH2-CH2-CH=N.

Explanation:

This is a spectroscopy question. Spectroscopy is an important part of chemistry which is used in the identification of chemical compounds.

So, without mincing words let's dive straight into the solution to the question.

Starting from values of IR given, the absorption at 1620-1680 cm^-1 shows the presence of C = N.

From the question, it is given that there is no signal showing on the spectrum at 3010 - 3090 cm^-1. The absorption at 3010 - 3090 cm^-1 shows the presence of N - H. But, it is not show on the spectrum. Hence, there is no N - H in the compound. That's, there are no Hydrogen atoms attached to Nitrogen atom.

We are given that the mass spectrum of compound A shows the molecular ion at m/z 85, therefore the molar mass of the compound is 85.

Also, from the question the elements present are carbon, Hydrogen and Nitrogen. Therefore, the compound is likely to be C5H11N.

The likely compounds are given below;

(1). CH3-CH=N-CH2-CH2-CH3.

(2). CH3-CH2-CH =N-CH2-CH3.

(3). CH3-CH2-CH2-CH=N-CH3.

(3). CH3-CH2-CH2-CH2-CH=N.

Given that a 0.130 M HCl(aq) solution costs $39.95 for 500 mL, and that KCl costs $10/ton, which analysis procedure is more cost-effective

Answers

Answer:

KCl is cost effective

Explanation:

In order to know this, we need to see how much it cost 1 g of each reactant. Let's begin with HCl

HCl:

In this case, let's calculate the moles of HCl in a 0.130 M solution and then, the mass of HCl using the molecular weight of 36.5 g/mol, to get the cost the HCl at the end using the given price:

nHCl = 0.130 moles/L * 0.5 L = 0.065 moles

mHCl = 0.065 moles * 36.5 g/mol = 2.3725 g

Cost HCl = 39.95 $ / 2.3725 g = 16.84 $/g

Conclusion, 1 g of HCl costs 16.84 $

KCl:

In this case, it's pretty obvious that 1 ton of KCl cost 10$, so, there is no need to do further calculations because 1 ton (or more than 1000 kg of the salt) it's just 10$. This is less expensive than the 16.84$ for just 1 g of HCl, so, final conclusion, KCl is more cost-effective.

Hope this helps

what are the characteristics of elastic energy

Answers

Elastic or potential energy is stored energy waiting to be released as kinetic energy. To be concise, stored energy.

The bonds between the atoms of an item or substance store elastic energy. When these bonds are under stress, they absorb energy and release it when they are relaxed.

What is elastic energy ?

Elastic energy is the mechanical potential energy held in a material's or a physical system's configuration while it is subjected to elastic deformation as a result of work being done on it.

Elastic energy is created when things are momentarily crushed, stretched, or otherwise distorted. Calculations of mechanical equilibrium locations using the elastic potential energy equation.

The energy is potential because when the item is permitted to return to its original shape by its elasticity, it will change into other types of energy, such as kinetic energy and sound energy.

Thus, The bonds between the atoms of an item or substance store elastic energy.

To learn more about elastic energy, follow the link;

https://brainly.com/question/14687790

#SPJ2

which metals take part in chemical reactions when exposed to the air?

Answers

Answer:

sodium metal take part when exposed to air

Soda pop is carbonated with CO2. Mark puts one bottle of soda pop in the refrigerator and leaves the other out in the hot sunlight. After one hour, he opens both bottles. Which bottle will likely have more fizzing and bubbles? Why?

Answers

Answer:

The one left in the hot sunlight.

Explanation:

The solubility of gases decreases when temperature increases. The gas in the soda pop (CO2) left in the sun will not stay dissolved as much as the on left in the refrigerator.

Soda is a carbonated drink with soda or water and a sweetener. Soda pop in the fridge will have more fizzing and bubbles.

What is carbonated?

Carbonated drinks are drinks or beverages that contain dissolved carbon dioxide. They have small bubbles and fizz of the carbon dioxide present in them.

Carbonation of the drinks is linked to the temperature as at high temperature, liquid tends to lose the carbon dioxide fizz. The carbon dioxide bubbles are retained at low temperatures as the solubility of the carbon dioxide is increased.

Therefore, the bottle kept in the refrigerator will have more bubbles and fizz.

Learn more about soda and carbonated here:

https://brainly.com/question/9908595

Define these terms: (a) unimolecular reaction (b) bimolecular reaction (c) elementary reaction (d) overall reaction

Answers

Answer:

See explanation

Explanation:

A unimolecular reaction is an elementary reaction in which the rearrangement of a single molecule produces one or more molecules of product(Chem Purdue).

A bimolecular reaction refers to the chemical combination of two molecular entities in a reaction that can be considered either reversible or irreversible(Springer).

Elementary reaction refers to any reaction that takes place in a single reactive encounter.

Overall reaction is the sum of all the steps of a non-elementary reaction.

Why is it important to keep detailed records and notes about evidence and the crime scene

Answers

Answer:

It is important to record the crime scene in order to view how it was originally found so when items are removed, it can still be investigated and viewed.

Explanation:

the pH of a solution prepared by the addition of 100mL 0.002M HCL to 100mL distilled water is closest to:
a.1.0
b.1.5
c.2.0
d.3.0

Answers

Answer:

d.3.0

Explanation:

Step 1: Calculate the final volume of the solution

The final volume is equal to the sum of the volumes of the initial HCl solution and the volume of distilled water.

V₂ = 100 mL + 100 mL = 200 mL

Step 2: Calculate the final concentration of HCl

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂ = 0.002 M × 100 mL/200 mL = 0.001 M

Step 3: Calculate the pH of the final HCl solution

Since HCl is a strong acid, [H⁺] = HCl. We will use the definition of pH.

pH = -log [H⁺] = -log 0.001 = 3

Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of metals A and B are 6.17 and 8.00 g/cm3 , respectively, and their respective atomic weights are 171.3 and 162.0 g/mol, determine whether the crystal structure for this alloy is simple cubic, facecentered cubic, or body-centered cubic. Assume a unit cell edge length of 0.332 nm

Answers

Answer:

Simple cubic

Explanation:

The density of metal A (ρa) = 6.17 g/cm³, The density of metal B (ρb) = 8 g/cm³, The atomic weight of metal A (Aa) = 171.3 g/mol, The atomic weight of metal B (Ab) = 162 g/mol, the unit cell edge length (a) = 0.332 nm, concentration of metal A (Ca) = 25%, concentration of metal B (Cb) = 75%

The average density is given by:

[tex]\rho_{ave}=\frac{100}{\frac{C_a}{\rho_a} +\frac{C_b}{\rho_b} } \\\\\rho_{ave}=\frac{100}{\frac{25}{6.17} +\frac{75}{8} } =7.45\ g/cm^3\\\\The\ average\ atomic\ weight\ is:\\\\A_{ave}=\frac{100}{\frac{C_a}{A_a} +\frac{C_b}{A_b} } \\\\A_{ave}=\frac{100}{\frac{25}{171.3} +\frac{75}{162} } =164.23\ g/mol\\\\The\ number\ of\ atoms\ per\ unit(n)\ is:\\\\n=\frac{\rho_{ave}*a^3*N_A}{A_{ave}} \\\\N_A=Avogadro\ constant=6.02*10^{22} \ mol^{-1},a=0.332\ nm=3.32*10^{-8}cm\\\\Substituting:\\\\[/tex]

[tex]n=\frac{\rho_{ave}*a^3*N_A}{A_{ave}} =\frac{7.45*(3.32*10^{-8})^3*6.02*10^{23}}{164.23} \\\\n=0.999\\[/tex]

n≅1

Since n≅1, the crystal structure for this alloy is simple cubic

Convert 65.4 m to mm.
Helppp please

Answers

Answer:65.4 meters= 65400 millimeters

65.4m = 65400mm.
Hope this helps

WHAT IS A PLACE WHERE BOOKS ARE CLASSIFIED! I NEED HELP!

Answers

Answer:

From the Online Catalog to the Shelf

Libraries in the United States generally use either the Library of Congress Classification System (LC) or the Dewey Decimal Classification System to organize their books. Most academic libraries use LC, and most public libraries and K-12 school libraries use Dewey.

Explanation:

Answer:

From the Online Catalog to the Shelf

Libraries in the United States generally use either the Library of Congress Classification System (LC) or the Dewey Decimal Classification System to organize their books. Most academic libraries use LC, and most public libraries and K-12 school libraries use Dewey.

You are working with a concentrated solution of ammonium hydroxide which place of safety equipment is most important to have on hand

Answers

I would say safety goggles
Lab coat, eye protection and gloves

Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. Include hydrogen atoms, nonbonding electrons, and formal charge(s) in your structure. The starting alkene is a 4 carbon chain with a double bond between carbons 2 and 3. The substituents on the alkene are on opposite sides of the alkene. This reacts with B r 2 to give the intermediate ion.

Answers

Answer:

See explanation and image attached

Explanation:

The reaction of bromine molecule with an alkene passes through a bridged intermediate known as the brominium ion.

It is a cyclic intermediate that contains a positively charged bromine ion as i have shown in the image attached.

The brominium ion is first formed during the bromininaton of alkenes.

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