The formal charges of all nonhydrogen atoms are -1.
Solution:-
<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>
O-Cl-O 6-7 = -1x4 = -4 N 5-4=1 H-N-H 1-1=0
O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1
<u>6-6 = 0 1-2 = -1</u>
It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.
According to the lewis structure of SO2, The critical atom is sulfur and it is bonded with 2 oxygen atoms thru a double bond. each oxygen atom acquires 2 lone pairs of electrons and the primary sulfur atom has 1 lone pair of electrons.
Learn more about Nonhydrogen atoms here:-brainly.com/question/2822744
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Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
S2F10, FE3N2, S7O2, NaCO3, Al2O3, PC13, and CO2. Hope this helps
<span>1)false a in chemical equilibrium concentration of reactant is equal to concentration of product
2)as here they said heat is added in product side means its endothermic reaction and in endothermic reaction on increasing temp. equilibrium shift towards forward direction so its true
3) B)as here mole are equal in reactant and product side that is 2 and if we increase pressure equilibrium shift in dat direction where no. of moles are less and here mole are equal so it will remain unaffected</span>
Answer:I belive it would be attracted seeing as how there are more magmatic charges on that side of no 2 and how there are more positive charges on the middle side of ballon no1.
Explanation:
