Answer:
Mo(CO)5 is the intermediate in this reaction mechanism.
Explanation:
The reaction mechanism describes the sequence of elementary reactions that must occur to go from reactants to products. Reaction intermediates are formed in one step and then consumed in a later step of the reaction mechanism.
In this reaction mechanism, Mo(CO)5 is the product of 1st reaction and then it is used as a reactant in 2nd reaction. So, Mo(CO)5 is the reaction intermediates.
The overall balanced equation would be,
Mo(CO)6 + P(CH3) ↔ CO + Mo(CO)5 + P(CH3)3
Answer: True
Explanation:
Let's begin by explaining that the vision phenomenon depends on three elements: the observer, the object and the light source.
In addition, objects have pigments, which generally absorb more light than they reflect (they absorb certain wavelengths and reflect others). Therefore, the color that a given object seems to have depends on which parts of the visible electromagnetic spectrum are reflected and which parts are absorbed.
In this sense, the colours we see are in fact the wavelengths that are reflected or transmitted from the object. For example, a red object has that color because when it is illuminated whith white light, the pigments on this object abrsorb all the the wavelengths of the visible electromagnetic spectrum, except red. That is why red light is the only light that is reflected from the mentioned object.
X=0.031903 I think if you don’t know how to do this photo math would be a good thing for you
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
Answer:
Lithium does form a peroxide as well as an oxide on burning in air and I suspect the low temperature reaction with air forms a significant amount of peroxide.
