Answer:
The electron pair geometry is Trigonal planar
Molecular geometry - Bent
Approximate bond angle - <120°
Explanation:
The valence shell electron pair repulsion theory enables us to predict the shapes of molecules based on the number of electron pairs present on the valence shell of the central atom and based on the hybridization state of the central atom.
sp2 hybridization corresponds to trigonal planar geometry. Let us recall that the presence of lone pairs causes a deviation of the molecular geometry from the expected geometry based on the number of electron pairs.
Hence, owing to one lone pair present, the observed molecular geometry is bent.
Answer:
They are the smallest units of matter
Explanation:
Atom are the basic units of a chemical element; the smallest unit of an element that retains the properties of an element
Answer: There is 125 mL of a 2.0 M CuCl2 solution are needed to make 500 mL of a 0.5 M solution
Explanation:
Given:
= 2.0 M,
= ?
= 0.5 M,
= 500 mL
Formula used to calculate the volume is as follows.
![M_{1}V_{1} = M_{2}V_{2}](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D)
Substitute the values into above formula as follows.
![M_{1}V_{1} = M_{2}V_{2}\\2.0 M \times V_{1} = 0.5 M \times 500 mL\\V_{1} = 125 mL](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D%5C%5C2.0%20M%20%5Ctimes%20V_%7B1%7D%20%3D%200.5%20M%20%5Ctimes%20500%20mL%5C%5CV_%7B1%7D%20%3D%20125%20mL)
Thus, we can conclude that there is 125 mL of a 2.0 M
solution are needed to make 500 mL of a 0.5 M solution.
Answer:
New predators and not a lot of food
Explanation:
If new predators come than they can eat the insects and go extinct. Also if there is not a lot if food than the will die because they have nothing to eat.
Answer:
![m_{O_2}=87.2gO_2](https://tex.z-dn.net/?f=m_%7BO_2%7D%3D87.2gO_2)
Explanation:
Hello.
In this case, given the chemical reaction, we can compute the grams of oxygen by using the 98.2 g of water via the 2:1 mole ratio between them, the molar mass of water that is 18.02 g/mol, the molar mass of gaseous oxygen that is 32.00 g/mol and the following stoichiometric procedure relating the given information:
![m_{O_2}=98.2gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molO_2}{2molH_2O}*\frac{32.00gO_2}{1molO_2} \\\\m_{O_2}=87.2gO_2](https://tex.z-dn.net/?f=m_%7BO_2%7D%3D98.2gH_2O%2A%5Cfrac%7B1molH_2O%7D%7B18.02gH_2O%7D%2A%5Cfrac%7B1molO_2%7D%7B2molH_2O%7D%2A%5Cfrac%7B32.00gO_2%7D%7B1molO_2%7D%20%20%20%5C%5C%5C%5Cm_%7BO_2%7D%3D87.2gO_2)
In which the result is displayed with three significant figures because the given mass of water 98.2 g, has three significant figures too.
Best regards!