Which of the following is not an example of a molecule

dangina [55]

Democritus was the first to propose the idea of the atom. He said the atom was just this tiny, solid sphere. However, he used no scientific evidence to support his claim, so a guy named John Dalton did some experimenting and basically backed up Democritus' claim with evidence. Then, a guy named J.J. Thompson came along and said the atom was not solid and that is consisted of tiny negatively charged particles(electrons) and he came up with the Plum Pudding model which is just a tiny sphere with a punch of random scattered dots in it. After that, Ernest Rutherford did experiments and found that the tiny sphere is made up of mostly empty space with a tiny, dense, positively charged sphere inside of it, and the negatively charged particles just randomly float around it. Neils Bohr then said that the electrons take specific, circular, evenly spaced paths. Then, finally, we come to the Quantum Mechanical Model which is the one accepted today. This model basically vetos Bohr's idea and has a nucleus inside of an electron cloud, which is where the electrons are found.

6 0

3 years ago

5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?

sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's Law</h3>

Given Data

number of moles of Ne = 5.1169 mol

Pressure = 0.9148 atm

Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

V = 418.14 L

Learn more about Boyle's law here:

brainly.com/question/469270

5 0

2 years ago

A stable atom that has a large nucleus most likely contains

maria [59]

Answer:

3. Equal numbers of protons and neutrons

Explanation:

The nucleus becomes unstable if the ratio of protons to neutrons is less than 1:1 or more than 1:1.5.

The most stable nucleus has a neutron proton ratio of 1:1 which means that they can not release a neutron or a proton to decay.

Nucleus 3 is therefore the most stable.

8 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

What is a huge organic molecule composed of many smaller molecules called?

photoshop1234 [79]

Answer:

Macromolecules. A very large organic molecule composed of many smaller molecules, 1)Carbohydrates, 2)proteins, 3)lipids, 4)nucleic acids. Three of the four classes of macromolecules that are polymers. 1.Carbohydrates.

7 0

3 years ago