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gtnhenbr [62]
3 years ago
6

What is the molecular structure of water? What are the physical and chemical properties of water?

Chemistry
1 answer:
Slav-nsk [51]3 years ago
6 0
Water (H
2O) is a polar inorganic compound that is at room temperature a tasteless and odorless liquid, which is nearly colorless apart from an inherent hint of blue. It is by far the most studied chemical compound and is described as the "universal solvent" [18][19] and the "solvent of life".[20] It is the most abundant substance on Earth[21] and the only common substance to exist as a solid, liquid, and gas on Earth's surface.[22] It is also the third most abundant molecule in the universe.[21]

Water (H
2O)





NamesIUPAC name

water, oxidane

Other names

Hydrogen hydroxide (HH or HOH), hydrogen oxide, dihydrogen monoxide (DHMO) (systematic name[1]), hydrogen monoxide, dihydrogen oxide, hydric acid, hydrohydroxic acid, hydroxic acid, hydrol,[2] μ-oxido dihydrogen

Identifiers

CAS Number

7732-18-5 

3D model (JSmol)

Interactive image

Beilstein Reference

3587155ChEBI

CHEBI:15377 

ChEMBL

ChEMBL1098659 

ChemSpider

937 

Gmelin Reference

117

PubChem CID

962

RTECS numberZC0110000UNII

059QF0KO0R 

InChI

InChI=1S/H2O/h1H2 

Key: XLYOFNOQVPJJNP-UHFFFAOYSA-N 

SMILES

O

Properties

Chemical formula

H
2OMolar mass18.01528(33) g/molAppearanceWhite crystalline solid, almost colorless liquid with a hint of blue, colorless gas[3]OdorNoneDensityLiquid:[4]
0.9998396 g/mL at 0 °C
0.9970474 g/mL at 25 °C
0.961893 g/mL at 95 °C
Solid:[5]
0.9167 g/ml at 0 °CMelting point0.00 °C (32.00 °F; 273.15 K) [a]Boiling point99.98 °C (211.96 °F; 373.13 K) [6][a]SolubilityPoorly soluble in haloalkanes, aliphaticand aromatic hydrocarbons, ethers.[7]Improved solubility in carboxylates, alcohols, ketones, amines. Miscible with methanol, ethanol, propanol, isopropanol, acetone, glycerol, 1,4-dioxane, tetrahydrofuran, sulfolane, acetaldehyde, dimethylformamide, dimethoxyethane, dimethyl sulfoxide, acetonitrile. Partially miscible with Diethyl ether, Methyl Ethyl Ketone, Dichloromethane, Ethyl Acetate, Bromine.Vapor pressure3.1690 kilopascals or 0.031276 atm[8]Acidity (pKa)13.995[9][10][b]Basicity (pKb)13.995Conjugate acidHydroniumConjugate baseHydroxideThermal conductivity0.6065 W/(m·K)[13]

Refractive index (nD)

1.3330 (20 °C)[14]Viscosity0.890 cP[15]Structure

Crystal structure

Hexagonal

Point group

C2v

Molecular shape

Bent

Dipole moment

1.8546 D[16]Thermochemistry

Heat capacity (C)

75.375 ± 0.05 J/(mol·K)[17]

Std molar
entropy (So298)

69.95 ± 0.03 J/(mol·K)[17]

Std enthalpy of
formation (ΔfHo298)

−285.83 ± 0.04 kJ/mol[7][17]

Gibbs free energy (ΔfG˚)

−237.24 kJ/mol[7]
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412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
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Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

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412.072 g Cl₂ ≈ 412 g Cl₂

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Explanation:

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Considering the Boyle's law, as the pressure decreases, volume increases and has a value of 2.246 mL.

<h3>Boyle's law</h3>

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

This law says that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

If an initial state 1 and a final state 2 are analyzed, Boyle's law is expressed as:

P1×V1=P2×V2

<h3>Volume at the surface of the lake</h3>

In this case, you know:

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Replacing in Boyle's law:

3.46 atm× 0.650 mL= 1 atm×V2

Solving:

V2= (3.46 atm× 0.650 mL)÷ 1 atm

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Finally, as the pressure decreases, volume increases and has a value of 2.246 mL.

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