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pentagon [3]
3 years ago
10

Look at the atom models you made in today’s activities. Name the three parts of the atom. A. protons, neutrons, and electrons B.

atoms, molecules, and nucleus C. protons, neutrons, and elements D. protons, molecules, and electrons
Chemistry
1 answer:
Musya8 [376]3 years ago
5 0
D, protons,molecules, and electrons
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With high cholesterol, you can develop fatty deposits in your blood vessels. Eventually, these deposits grow, making it difficult for enough blood to flow through your arteries. Sometimes, those deposits can break suddenly and form a clot that causes a heart attack or stroke.

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For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.50
ira [324]

Answer:

The van't hoff factor of 0.500m K₂SO₄ will be highest.

Explanation:

Van't Hoff factor was introduced for better understanding of colligative property of a solution.

By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.

a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.

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3 years ago
Which is the correct Lewis dot structure of NH2?
Svetlanka [38]

Answer:

D

OE. E

Explanation:

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3 years ago
What is an atomic number?
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An atomic number is <span>the number of protons in the nucleus of an atom, which determines the chemical properties of an element and its place in the periodic table or chart.</span>
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The steps required to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25x10^-2 M. Please
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Answer:

In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.

Explanation:

Concentration: 1.25 x 10⁻² M

1,25 x 10⁻² mol FeCl₃ ___ 1000 mL

              x                   ___ 200.0 mL

         x = 2.5 x 10⁻³ mol FeCl₃

Mass of FeCl₃:

1 mol FeCl₃ _____________ 162.2 g

2.5 x 10⁻³ mol FeCl₃ _______    y

                  y = 0.4055 g FeCl₃

8 0
4 years ago
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