Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.
The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.
<h3>What is the Henderson-Hasselbalch equation?</h3>
Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.
Given,
Acid disassociation constant (ka) = 1. 8 10⁻⁵
Concentration of NaOH = 2.0 M
Concentration of CH₃COOH = 2.0 M
pKa value is calculated as,
pKa = -log Ka
pKa = - log (1. 8 x 10⁻⁵)
Substituting the value of pKa in the Henderson-Hasselbalch equation as
pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]
pH = - log (1. 8 x 10⁻⁵) + log [1]
= 4.745 + 0
= 4.75
Therefore, 4.75 is the pH of the buffer.
Learn more about the Henderson-Hasselbalch equation here:
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Answer:
C: 4
H: 1
Mg: 2
O : 6
Explanation:
You can quickly find out the number of valence electrons by looking at where the element is on the periodic table and referring to the table that is attached.
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
