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3 years ago

Name two physical properties that characterize matter.

Chemistry

2 answers:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

Physical properties are used to observe and describe matter. Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others.

Explanation:

KATRIN_1 [288]3 years ago

5 0

Answer:

Matter is anything that has mass and occupies a space. Matter has 2 properties physical properties and chemical properties. And 2 physical properties include mass and shape (solid, liquid and gas).

Other physical properties of molecular matter include:

Volume

Color

Odor

Luster

Hardness

Melting Point

Freezing Point

Boiling Point

Density

Malleability

Ductility

Conductivity

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In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 18.0°C. You identified that the ideal pressure (

Natasha2012 [34]

Answer:

4,38%

small molecular volumes

Decrease

Explanation:

The percent difference between the ideal and real gas is:

(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ 4,38%

This difference is considered significant, and is best explained because argon atoms have relatively small molecular volumes. That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.

Therefore, an increasing in volume will produce an ideal gas behavior. Thus:

If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to decrease

I hope it helps!

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3 years ago

Choose the thermochemical equation that illustrates ΔH°f for Li2SO4. Choose the thermochemical equation that illustrates ΔH°f fo

Deffense [45]

Answer:

2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)

Explanation:

A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.

The only equation that meets those conditions is the last one.

A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.

C is wrong because Li⁺ and SO₄²⁻ are not elements.

D is wrong because it shows the formation of 8 mol of Li₂SO₄.

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2 years ago

Which group is the most reactive?

RideAnS [48]

Answer:

alkali metals- Group 1

Explanation:

they have less valence electrons and therefore are more reactive

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3 years ago

प्रश्न 5. 'पुष्प की अभिलाषा' कविता में पुष्प के दवारा क्या अभिलाषा व्यक्त की गई

Kay [80]

Please ask in English

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2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

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2 years ago