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Marina86 [1]
2 years ago
13

Which items below are properties of matter? [select all that apply]

Chemistry
1 answer:
8_murik_8 [283]2 years ago
4 0

Answer:

Solubility, length, hardness, color, mass, density, weight, volume, boiling, and point.

Explanation: Hope this helps

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Chuẩn Độ 15 ml dung dịch CH3COOH 0,2 m bằng dung dịch NaOH 0,2 m
mina [271]

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Explanation:

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3 0
3 years ago
How many significant figures are in the measurement 0.03050
faltersainse [42]
There are 4 significant figures(3050)
8 0
3 years ago
Which of the following factors affects the strength of the gravitational force between two objects?
mash [69]
D. the distance between the objects
6 0
3 years ago
Read 2 more answers
In which state of the following compounds does nitrogen have the most positive oxidation state? group of answer choices no2 nano
Amiraneli [1.4K]

The compound nitrogen have most positive oxidation state is NO₂. The correct option is b.

<h3>What is oxidation state?</h3>

The total number of electrons gained or lost by an atom in order to form a chemical bond with another atom.

The charge on an ion is equal to the sum of the oxidation states of all the atoms in the ion. A substance's more electronegative elements are given a negative oxidation state.

A positive oxidation state is assigned to the less electronegative element.

Thus, the correct option is b, NO₂.

Learn more about oxidation state

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4 0
2 years ago
635 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at standard pressure (STP)?
Sveta_85 [38]

The new volume at standard temperature and pressure is 5.08 L.

Explanation:

As per the kinetic theory of gases, the volume occupied by gas molecules will be inversely proportional to the pressure of the gas molecules. This is termed as Boyle's law.

So, pressure∝\frac{1}{volume}

Thus, if two pressure and two volumes are given then,

P_{1} V_{1} = P_{2} V_{2}\\

Now, we known the values of P₁ = 8 atm, V₁ = 635 mL, P₂ = 1 atm and V₂ we have to determine. We are considering P₂ = 1 atm, because we have to determine V₂ at standard temperature and pressure. And standard pressure is 1 atm.

8*635*10^{-3}  = 1 *V{2} \\\\V_{2} = \frac{8*635*10^{-3} }{1} =   5.08 L

Thus, the new volume at standard temperature and pressure is 5.08 L.

5 0
3 years ago
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