<u>Answer:</u> The number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
<u>Explanation:</u>
In a chemical equation, the chemical species are termed as reactants or products.
Reactants are defined as the species which react in the reaction and are written on the left side of the reaction arrow.
Products are defined as the species which are produced in the reaction and are written on the right side of the reaction arrow.
For the given chemical equation:
On the reactant side:
Number of carbon atoms = (6 × 2) = 12
Number of hydrogen atoms = (14 × 2) = 28
Number of oxygen atoms = (2 × 19) = 38
Hence, the number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
<span>The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to </span>extracellular fluid and equilibrium is reached.
Answer:
C
Explanation:
Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.
A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.
Answer:
V¹N²= V²N²
here V¹= ?
N¹= 6.00
V²= 175ml
M²= 0.2M
So V¹= (V²N²)/N² = (175 x 0.2)/6
V¹ = 5.83 ml
Explanation:
Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.
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.
