Explanation:
1)
Mass of NaOH = m
MOlar mass of NaOH = 40 g/mol
Volume of NaOH solution = 1.00 L
Molarity of the solution= 1.00 M
A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.
Upto two significant figures mass should be determined.
2)
(dilution equation)
Molarity of the NaOH solution = 
Volume of the solution = 
Molarity of the NaOH solution after dilution = 
Volume of NaOH solution after dilution= 
A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.
Upto three significant figures volume should be determined.
Answer:
Explanation:
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a highly unreactive carboxylate. However when the ammonium carboxylate salt is heated to a temperature above 100 C water is driven off and an amide is formed.
Answer:
V = 22.42 L/mol
N₂ and H₂ Same molar Volume at STP
Explanation:
Data Given:
molar volume of N₂ at STP = 22.42 L/mol
Calculation of molar volume of N₂ at STP = ?
Comparison of molar volume of H₂ and N₂ = ?
Solution:
Molar Volume of Gas:
The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol
Molar volume can be calculated by using ideal gas formula
PV = nRT
Rearrange the equation for Volume
V = nRT / P . . . . . . . . . (1)
where
P = pressure
V = Volume
T= Temperature
n = Number of moles
R = ideal gas constant
Standard values
P = 1 atm
T = 273 K
n = 1 mole
R = 0.08206 L.atm / mol. K
Now put the value in formula (1) to calculate volume for 1 mole of N₂
V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm
V = 22.42 L/mol
Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.
Answer:
The H in the carboxyl group.
Explanation:
Acetic acid can be written as CH₃COOH, where -COOH is the functional group carboxyl, responsible for the acidity of organic acids. The H in the carboxyl group is the one that is donated in the acid reaction.
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺
Acetic acid is a weak acid, so just a small fraction of the molecules undergo this reaction to donate their hydrogen.
From the calculations, the concentration of the acid is 0.24 M.
<h3>What is neutralization?</h3>
The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.
We have to use the formula;
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O
CA = ?
CB = 1.2 M
VA = 50 mL
VB = 20 mL
NA = 1
NB = 2
CA = CBVBNA/VANB
CA = 1.2 M * 20 mL * 1/ 50 mL * 2
CA = 0.24 M
Learn more about neutralization:brainly.com/question/27891712
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