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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The electrons are unequally shared. The electronegative element receives the electrons from the electropositive one.
The correct answer is option B. The most dense phase of matter is the solid phase and the least dense are gases. However, there is an exception. Water is the exception. Solid water or ice is less dense than the liquid phase therefore it floats on liquid water.
Answer:
Explanation:
We must do the conversions
mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O
m/g: 24.5
(a) Moles of C₆H₁₂O₆
(b) Moles of CO₂
(c) Volume of CO₂
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.960 atm
n = 0.8159 mol
T = 37 °C
(i) Convert the temperature to kelvins
T = (37 + 273.15) K= 310.15 K
(ii) Calculate the volume
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
