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Lelu [443]
1 year ago
9

Ice has a density of 0.92g/cm3. It will float in water.

Chemistry
1 answer:
SVEN [57.7K]1 year ago
6 0

The density of pure water is 1 g/cm^3.

Its density is 0.98 g cm 3 at room temperature, in comparison with the handiest zero.92 g cm 3 for ice, a reality that has to be defined through atomic, and molecular concepts. If ice has been no longer much less dense than water, it might sink, having a devastating impact on lake backside ecosystems. believe it or now not, ice is honestly about 9% much less dense than water. for the reason that water is heavier, it displaces the lighter ice, causing the ice to glide to the pinnacle.

The density of ice is about 90 percent that of water, but that could range because ice can contain air, too. meaning that about 10 percent of an ice cube or iceberg will be above the water line. The density of water is maximum at four∘C, and the density of the ice is much less than the water due to its susceptible intermolecular pressure of attraction. as the density of water is more, it's miles heavier than ice. therefore ice floats on the floor of the water. Ice continually floats due to the fact it's far less dense than everyday water. because frozen water molecules shape a crystal.

Learn more about A density here:-brainly.com/question/17780219

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As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is
chubhunter [2.5K]

Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
Precipitation calculations with Ni2+ and Pb2+
Maurinko [17]

Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

When Ni(OH)₂ starts precipitate :

Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]

5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]

[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18

[ OH⁻ ] = 5.5 × 10⁻⁸ M

pOH = 7.2

therefore , pH = 14 - 7.2

                  pH = 6.8

Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

To learn more about pH here

brainly.com/question/15289741

#SPJ1

6 0
1 year ago
A chemical reaction at equilibrium always has _______.
natka813 [3]

Answer:

A

Explanation:

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3 years ago
The effect of nitogen and plants. Which factor is an independent variable
Yakvenalex [24]
The independent variable is what you are testing in the experiment - it sounds like the independent variable in this experiment is the amount/concentration of nitrogen
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denpristay [2]
Indicators are colorful compounds that are weak acids or bases.
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