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3 years ago

Classify each property as intensive or extensive. Color, temperature, volume, hardness, mass, boiling point, length, density

Chemistry

2 answers:

lisov135 [29]3 years ago

8 0

Explanation:

When amount of matter changes in a substance then a physical property that does not change due to this is known as an intensive property.

For example, density, color, boiling point, hardness etc are all intensive properties.

Whereas property of a substance that changes due to change in amount of matter of a substance is known as extensive property.

For example, temperature, volume etc are all extensive properties.

Hence, we can conclude that given properties are classifies as follows.

Intensive property : color, hardness, boiling point, density, temperature.

Extensive property : length, volume, mass.

mr_godi [17]3 years ago

3 0

Answer:

Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.

Explanation:

Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:

Boiling Point

Density

State of Matter

Color

Melting Point

Odor

Temperature

Refractive Index

Luster

Hardness

Ductility

Malleability

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Order of metals from least reactive to most reactive: B <C <A <D

<h3>Further explanation</h3>

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction

Let's analyze the statement in the problem

I. Only A, C and D react with 1 mol/L HCl to give H₂(e)

M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)

A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H

II. When A is added to solutions of the other metal ions, metallic B and C are formed but not D.

This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive

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3 years ago

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Karo-lina-s [1.5K]

Answer:

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3 years ago

A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th

Kisachek [45]

Answer:

$pH=4.1$

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

$pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1$

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4 years ago

Question 5 of 10

djyliett [7]

A. A diagram showing the effects of temperature and pressure on phase

Explanation:

A phase diagram is a diagram that shows the effects of temperature and pressure on phase.

A phase diagram shows how different substances are transformed from one form to another.
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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

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