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]
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped :D
Answer:
The reaction will shift in the direction of products.
Explanation:
<u>Step 1:</u> Data given
A reaction mixture contains:
0.41 M SO2
0.14 M NO2
0.12 M SO3
0.14 M NO
<u>Step 2:</u> The balanced equation
O2(g) + NO2(g) ↔ SO3(g) + NO(g) Kc = 0.33
<u>Step 3:</u> Define the direction of the shift of reaction:
When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.
When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.
When Q=K, the system is at equilibrium and there is no shift to either the left or the right.
<u>Step 4:</u> Calculate Q
Q = [NO][SO3]/[SO2][NO2]
Q = (0.14 *0.12)/(0.41*0.14)
Q = 0.0168/0.0574
Q = 0.293
Q<Kc
This means there are more reactants than products. Thud, some of the reactants will become products, causing the reaction to shift to the right.
The reaction will shift in the direction of products.
Answer:
Explanation:
Hello,
In this case, the reaction is:
Thus, the law of mass action turns out:
![Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_3CH_2OH%5D_%7Beq%7D%7D%7B%5BH_2O%5D_%7Beq%7D%5BCH_2CH_2%5D_%7Beq%7D%7D)
Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change 
result:
![[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol](https://tex.z-dn.net/?f=%5BCH_2CH_2%5D_%7Beq%7D%3D29mol-x%3D16mol%5C%5Cx%3D29-16%3D13mol)
In such a way, the equilibrium constant is then:
Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:
Thus, the second change, 
finally result (solving by solver or quadratic equation):
Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:
Best regards.
