Identify the element that cannot participate in nuclear fission reactions.

Which property of water explains why seawater contains over 20 different elements? select one:

Gnesinka [82]

I believe the awnser is a. universal solvent.

8 0

3 years ago

Read 2 more answers

Analyse the temperature versus time graph of water, given below.

BlackZzzverrR [31]

Answer:

Phase C - Liquid State

Phase E - Gaseous State

Explanation:

Usually, in phases of water, we have the following;

When temperature is less than zero, it is said to be in its solid phase as ice.

When temperature is between 0 to 100, we can say it is in the liquid phase as water.

When temperature is above 100°C, It is said to be in the gaseous phase as vapour.

From the diagram;

Phase C is the only liquid state because it falls between temperature of 0°C and 100°

Also, only phase E is in the gaseous phase because the temperature is above 100°C.

3 0

2 years ago

The ksp of yttrium fluoride, yf3, is 8.62 × 10-21. calculate the molar solubility of this compound.

motikmotik

Answer:

The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.

Explanation:

In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.

YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)

I 0 0

C +S +3S

E S 3S

The solubility product (Ksp) is:

Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴

$S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21} /27}=4.23 \times 10^{-6}M$

6 0

3 years ago

Swer questions 12 B D 12 Classify each model A-D as either an ele- ment, a compound, or a mixture. Explain your reasoning for ea

MrRissso [65]

Answer:

13-b

Explanation:

because 13-b is close to a and 12

3 0

2 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