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Amanda [17]
3 years ago
11

Name three pure elements and three pure compounds​

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

Pure elements:

Hydrogen(H)

Nitrogen(N)

Magnesium (Mg)

Pure Compounds:

Oxygen gas(O2)  

Water (H2O)  

Ammonia (NH3)

Explanation:

Hope it helps.Please mark me brainliest.^-^

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It would most likely be a observation or hypothesis
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Guy plzz help me the this and thanks
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Answer:

<h2>18 {cm}^{3}</h2>

Explanation:

<em><u>In</u></em><em><u> </u></em><em><u>first</u></em><em><u> </u></em><em><u>picture</u></em><em><u>,</u></em>

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<em><u>In</u></em><em><u> </u></em><em><u>second</u></em><em><u> </u></em><em><u>picture</u></em><em><u>,</u></em><em><u> </u></em>

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4 0
3 years ago
A mole of one element contains _______ number of atoms as a mole of another element.
Vera_Pavlovna [14]

A mole of one element contains Same number of atoms as a mole of another element.

The atomic mass of an element, that is found on the periodic table, may be used to determine the amount of moles in a system. Typically, this mass represents the average of the element's abundant forms found on earth. The mass of an element is given as the average of all of its earthly isotopes. The molar mass of a material is the weight of a mole of that substance. In chemistry, the molar mass is frequently used in converting grams of a chemical to moles. The periodic table lists an element's molar mass, which is its atomic weight in grams per mole (g/mol).

The average mass of an element's atoms expressed in atomic mass units is known as its atomic mass (amu, also known as daltons, D). The weight of the each isotope is combined by its abundance to get the atomic mass, which is a weighted average of all the isotopes of that element.

To know more about Moles refer to brainly.com/question/19638114

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1 year ago
What is the volume of 14.0 g of nitrogen gas, N2, at a pressure of 2.00 atm and temperature of 27.0 degree celsius
Lelu [443]
P = 2.00 atm = 2026.50hPa
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m = 14.0g              |
Mr = 28g/mol         |      n = m/M = 14g/28g/mol = 0,5mol
R = 83.1 [hPa*L]/[mol*K]

pV = nRT

        nRT        0,5mol * 83.1[hPa*L]/[mol*K] * 300K
V = ---------- = ----------------------------------------------------------- = <u>6.15L</u>
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