40 electrons
Explanation:
The N for silicon tetrachloride, SiCl₄ is 40 electrons. The needed electrons that would be used to complete the lewis structure of the compound is actually 40 electrons. This number of electron will help the compound attain a noble configuration.
- The compound SiCl₄ is a covalent one. Here, there is sharing of electrons between two atoms.
- In drawing the electron dot formula, one must take into account the Available electrons and the Needed electrons.
- The Available electrons are sum of the valence electrons that can be accessed for the bonding. Si has 4 valence electrons, Cl has 7 valence electrons this makes a total of 4 + 7(4 atoms of chlorine), 32 available electrons.
- But to make a complete octet like that of noble gases, each atom most have 8 complete outer most electrons. This is the needed number of electrons. Since there are 5 atoms i.e 4 atoms of chlorine and 1 atom of Silicon, the needed electrons will be 5x8 = 40 electrons.
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lewis structure : brainly.com/question/6215269
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-they are inside the nucleus of an atom
- they have a relative charge of +1
- they have a relative mass of 1
Answer:
two examples are blood and soapy water.
Explanation:
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
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