Answer:
True:
- Hybrid orbitals within the same atom have the same energy and shape.
- Hybrid orbitals are described mathematically as a linear combination of atomic orbitals.
- An atom can have both hybridized and unhybridized orbitals at the same time.
Explanation:
Hybrid orbitals do not exist in isolated atoms. They form only in covalently bonded atoms.
Hybridization happens when several atomic orbitals combine to form other orbitals with the same energy and greater stability.
A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that combined to produce the set.
Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds, and both can appear in an atom at the same time.
Answer:
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Explanation:
Average rate of the reaction is defined as ratio of change in concentration of reactant with respect to given interval of time.
![R_{avg}=-\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
Where :
= initial concentration of reactant at 
.
= Final concentration of reactant at 
.
2A+3B → 3C+2D
![R_{avg}=-\frac{1}{2}\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
The concentration of A at (
) = 
The concentration of A at (
) = 
The average rate of reaction in terms of the disappearance of reactant A in an interval of 0 seconds to 20 seconds is :
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
First, we apply the law of conservation of mass which states that the total mass in a system remains constant.
Therefore, there must be 5.00 g of sulfur and 4.99 g of oxygen in the product. Now, we determine the mass percentage using:
Mass % = (mass of sulfur x 100) / total mass of compound
Mass % = (5 * 100) / (5 + 4.99)
Mass % = 50.05%
The product contains 50.05% sulfur by mass.
Answer:
3.33 M
Explanation:
It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:
" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>
<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "
In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:
- 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃
Then we<u> convert the 150 mL of the volumetric flask into L</u>:
Finally we <u>divide the moles by the volume</u>:
- 0.500 mol AgNO₃ / 0.150 L = 3.33 M
