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Greeley [361]
2 years ago
5

Life as we know it could not exist without water. All the chemical reactions of life occur in an aqueous solution. Water molecul

es are polar and are capable of forming hydrogen bonds with other polar or charged molecules. As a result, water has the following properties:
A. H2O molecules are cohesive; they form hydrogen bonds with each other.
B. H2O molecules are adhesive; they form hydrogen bonds with polar surfaces.
C. Water is a liquid at normal physiological (or body) temperatures.
D. Water has a high specific heat.
E. Water has a high heat of vaporization (energy needed to evaporate)
F. Water's greatest density occurs at 4°C
Chemistry
1 answer:
dusya [7]2 years ago
5 0
A) At 0 C water forms ice but as mentioned above F) water's greatest density occurs at 4 C and it decreases below 4 C so ice is lighter than 4C water, thus, at 0C ice comes to surface and acts as insulator thereby preventing lower water from freezing.
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algol [13]
13. D
14. A
15. C
16. B
17. F
18. E
7 0
3 years ago
A sample of carbon monoxide gas is initially in a 5858 mL container. The gas is then moved to 3.29 L container at a temperature
Doss [256]

Answer:

d i took the test

Explanation:

5 0
3 years ago
Hydrogen is a reactive gas under normal conditions.Based on this property alone, to which group of the periodic table should hyd
denis23 [38]
Even though Hydrogen is originally in group 1, based on this property, we can say it is in group 6.

Because:

Group 6 would mean that it only needs 2 more valence electrons till the octet (8 valence electrons). This would make it reactive, yet, in normal conditions, unlike group 7.
5 0
4 years ago
Read 2 more answers
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
Tia has a sample of pure gold (Au). She weighed the sample and the result was 62.4 grams. Tia wants to determine the number of a
Marta_Voda [28]
I hope you are able to find your answer through the guidance of this site as I have yet to cover this topic in chemistry: http://scientifictutor.org/1021/chem-how-to-convert-between-grams-and-molecules/
7 0
3 years ago
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