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3 years ago

How is the number of valence electrons related to the ionization energy of an element?

1 answer:

8 0

A) Low ionization energy correlates to elements with fewer valence electrons.

Ionization energy is the energy required to separate valence electrons from their atom. The more valence electrons there are, the higher the ionization energy.

Hope this helps!

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Number of SO2 molecules in 1.28 mol of SO2

dolphi86 [110]

Answer:

7.7056 x 10^23

6 0

3 years ago

For each situation below, state whether water will move into the cell, move out of the cell, or stay the same.

Firdavs [7]

Water moves from an area of higher water potential (aka. "more water" in simple language) to an area of lower water potential (aka. "less water" in simple language).

For A, cells in carrots have water stored in their cytoplasm, where many soluble substances may be found (e.g. sodium ions). On the other hand, pure water has no other soluble substances other than the water molecules (I.e. H2O). Pure water will thus have a higher water potential as compared to the water in carrot cells, and so, water will move from pure water into the carrot cells via osmosis down a concentration gradient.

B. Corn syrup is water that has high concentrations of sugars, thus it is very likely to have a lower water potential than the cells of carrots. Water will move from within the cells of carrots and out to the corn syrup, down a concentration gradient.

C. The water in carrot cells will stay the same, since carrot cells have the same water potential as the surrounding solution which has the same water potential as cytoplasm.

Hope this helps! :)

8 0

3 years ago

When an ionic compound such as sodium chloride (NaCl) is placed in water, the component atoms of the NaCl crystal dissociate int

sesenic [268]

Answer:

The solution expected to contain the greatest number of solute particles is: A) 1 L of 1.0 M NaCl

Explanation:

The number of particles is calculated as:

a) For Ionic compounds:

molarity × volume in liters × number of ions per unit formula.

b) For covalent compounds:

molarity × volume in liters

The difference is a factor which is the number of particles resulting from the dissociation or ionization of one mole of the ionic compound.

So, calling M the molarity, you can write:

# of particles = M × liters × factor

This table show the calculations for the four solutions from the list of choices:

Compound kind Particles in solution Molarity # of particles

(dissociation) (M) in 1 liter

A) NaCl ionic ions Na⁺ and Cl⁻ 1.0 1.0 × 1 × 2 = 2

B) NaCl ionic ions Na⁺ anc Cl⁻ 0.5 0.5 × 1 × 2 = 1

C) Glucose covalent molecules 0.5 0.5 × 1 × 1 = 0.5

D) Glucose covalent molecules 1.0 1.0 × 1 × 1 = 1

Therefore, the rank in increasing number of particles is for the list of solutions given is: C < B = D < A, which means that the solution expected to contain the greatest number of solute particles is the solution A) 1 L of 1.0 M NaCl.

8 0

3 years ago

Solar system question

shtirl [24]

Answer:

yes

Explanation:

is that is the alteration of a planet on its orbit around the sun depends upon the mass of the Sun and the inverse square of the planets distance from the Sun. As a planet moves further away in its orbit around the Sun, the gravitational force exerted by the sun on the planet decreases

8 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