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JulijaS [17]
3 years ago
10

According to VSEPR theory, a molecule with the general formula AX2E2 (A is the central atom, X is the bonded atom and E is a lon

e pair of electrons) has how many regions of electron density around the central atom [ Select ]
Chemistry
1 answer:
ratelena [41]3 years ago
4 0

Answer:

see explanation

Explanation:

An AX₂E₂ geometry is derived from an AX₄ parent geometry and is based upon 4 regions of electron density about the central element and defines a tetrahedral geometry and the geometry is bent angular.

An example is the water molecule (H₂O) with two covalent O - H bonds and two free pair electrons on the central oxygen element.

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To solve for the number of moles, we simply have to use the Avogadros number which states that there are 6.022 x 10^23 molecules per mole. Therefore:
 
number of moles = 6.67 X 10^40 chlorine molecules / (6.022 x 10^23 molecules / mole)
number of moles = 1.108 x 10^17 moles



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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
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Answer:

m_{H_2O}=4.86gH_2O

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Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

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m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

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3 years ago
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