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3 years ago

Homogeneous mixtures

Chemistry

2 answers:

galina1969 [7]3 years ago

8 0

Explanation:

<h3>The mixture in which the particles of the components of solute and solvent are equally mixed is called homogeneous mixture. </h3>

Naily [24]3 years ago

6 0

A homogeneous mixture is simply any mixture that is uniform in composition throughout. as long as each substance is mixed in enough to be indistinguishable from the others, it is a homogeneous mixture.

examples: air, saline solution, most alloys, and bitumen

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Identify the following changes as physical or chemical.

andrezito [222]

Answer:

physical

Explanation:

5 0

3 years ago

Draw a correct Lewis structure for XeI2 (Xe in middle, surrounded by I's ) that puts a 0 formal charge on all atoms. How many lo

Alona [7]

1) The Lewis structure will be:

I--I

In this structure, the xenon atom is surrounded by two iodine atoms, which are bonded to it through single bonds. Each iodine atom has one lone pair of electrons, for a total of 2 lone pairs on the central atom (xenon).

2) The molar mass of the gaseous compound is 0.416 g/mol.

To draw a correct Lewis structure for XeI2, we need to first count the number of valence electrons in the molecule. Xenon is a noble gas and has 8 valence electrons, while iodine has 7 valence electrons for a total of 23 valence electrons. To satisfy the octet rule and put a 0 formal charge on all atoms, we can use the following Lewis structure:

I--I

To find the molar mass of the gaseous compound, we can use the ideal gas law:

M = dRT/P

Where M is the molar mass, d is the density, R is the ideal gas constant, T is the temperature, and P is the pressure.

Given that the density of the gas is 0.3876 grams/142 mL = 0.002736 grams/mL, the temperature is 150 + 273 = 423 K, the pressure is 775 torr = 775/760 atm = 1.0132 atm, and the ideal gas constant is 0.08206 L·atm/mol·K.

We can calculate the molar mass as follows:

M = (0.002736 g/mL) * (0.08206 L·atm/mol·K / (1.0132 atm)) * (423 K)

M = 0.416 g/mol

Learn more about molar mass, here brainly.com/question/12127540

#SPJ4

3 0

2 years ago

If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain afte

Assoli18 [71]

Answer:

$m=0.127g$

Explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

$k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}$

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

$m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g$

Best regards.

6 0

3 years ago

How do I do this? What are the answers to the 5 questions shown?

frozen [14]

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

6 0

3 years ago

URGENT PLEASE HELP!! Earth science!!

aev [14]

Answer:

URGENT PLEASE HELP!! Earth science!!

What stages of development are represented by the Group 2 and Group 3 stars? Describe these

groups in terms of their relative age, size, brightness, and temperature.

3 0

3 years ago