Answer: The correct answer is Option 1.
Explanation:
Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.
We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.
Hence, the correct answer is Option 1.
Answer:
Explanation:
The elements in Group I of the periodic table are called alkali metals. They are called alkali metals because they react with water to form alkali solutions. These metals are very reactive; hence they have to be stored under oil to protect them from corrosion by air and waterwaterwater
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]
Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
