The Lewis structure for H₂CO is shown in the attached picture. The central atom is the carbon. However, I'm not sure which bond you're referring to. There can be two answers. The two C-H bonds are sp³ hybridized because it is a single bond. The C=O bond is sp² hybridized because it is a double bond.
Answer
- continuous removal of PH3
- adding more of P into the system
Explanation:
In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);
- The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:
heat+A⇌BΔH=+
- Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
- According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
