Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
The correct answer is 2.53 g of precipitate, BaCrO4.
Answer:
Sucrose is a disaccharide composed of alpha D gluose and beta D fructose linked together by beta 2,alpha1 glycosidic linkage.
Explanation:
The specificity of glycosidic linkage very much essential to choose the substrate for the synthesis of specific disaccharide.
For example sucrose contain beta 2,alpha1 glycosidic linkage that means the hydroxyl group of anomeric carbon of one monosaccharide(fructose) should remain in beta conformation and the hydroxyl group of other monosaccharide(glucose) should remain in alpha conformation.
The complete question is
If the compound below is oxidized, the resulting product is ___.
the compound given is Butanal
Select one:
a. methane and propane
b. butanal
c. butanoic acid
d. butane
Answer:
C. Butanoic Acid
Explanation:
When Butanal is oxidized using reagents like KMnO4, Tollens reagent etc.
When Aldehyde is oxidized corresponding Carboxylic Acid is produced
Butanal → Butanoic Acid
