Since transition metals with partially filled d orbitals have color when in solution. Therefore which ever solution has some color will likely contain a transition metal with a partially filled d orbital.
an example of this is solution with Cu²⁺ will have a blue tint to it.
Answer : The complete question is attached in the answer.
Explanation : In the given question an alkene is treated with alkylhalide in presence of methylene dichloride it gives a corresponding alkylbromide species.
The mechanism follows markonikov's addition rule, and the major product thus formed is bromine which is attached to least substituted carbon atom ad minor is bromine attached to highly substituted carbon atom.
Procaryotic structural components consist of macromolecules such as DNA, RNA, proteins, polysaccharides, phospholipids, or some combination thereof. The macromolecules are made up of primary subunits such as nucleotides, amino acids and sugars (Table 1). It is the sequence in which the subunits are put together in the macromolecule, called the primary structure, that determines many of the properties that the macromolecule will have. Thus, the genetic code is determined by specific nuleotide base sequences in chromosomal DNA; the amino acid sequence in a protein determines the properties and function of the protein; and sequence of sugars in bacterial lipopolysaccharides determines unique cell wall properties for pathogens. The primary structure of a macromolecule will drive its function, and differences within the primary structure of biological macromolecules accounts for the immense diversity of life.
Answer is: 25.84 milliliters of sodium metal.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.
d(Na) = 0.97 g/mL; density of sodim.
m(NaOH) = 43.6 g; mass of sodium hydroxide.
n(NaOH) = m(NaOH) ÷ M(NaOH).
n(NaOH) = 43.6 g ÷ 40 g/mol.
n(NaOH) =1.09 mol; amount of sodium hydroxide.
From chemical reaction: n(NaOH) : n(Na) = 2 : 2 (1: 1).
n(Na) = 1.09 mol.
m(Na) = 1.09 mol · 23 g/mol.
m(Na) = 25.07 g; mass of sodium.
V(Na) = m(Na) ÷ d(Na).
V(Na) = 25.07 g ÷ 0.97 g/mL.
V(Na) = 25.84 mL.
Answer:
For the following reaction, 8.00 grams of silicon tetrafluoride are allowed to react with 7.40 grams of water. silicon tetrafluoride (s) + water (l) hydrofluoric acid (aq) + silicon dioxide (s) What is the maximum amount of hydrofluoric acid that can be formed? grams What is the FORMULA for the limiting reagent? H2O What amount of the excess reagent remains after the reaction is complete? grams