Answer:
2) 0.109 mol
Explanation:
The density of butan-1-ol is 0.8098g/mL. And its molar mass is: 74.121g/mol.
First, we need to convert volume of butan-1-ol to mass:
10.0mL * (0.8098g / mL) = 8.098g of butan-1-ol
Now, we need to convert these grams to moles using molar mass:
8.098g * (1mol / 74.121g) = 0.109 moles of butan-1-ol
Right answer is:
<h3>2) 0.109 mol
</h3>
On oxidation of aldehydes produces carboxylic acid functional group.
The product of oxidation of aldehydes depends upon whether the reaction occur in acidic medium or alkaline condition.
If oxidation of aldehydes occurs under acidic condition the product is carboxylic acid but if oxidation of aldehydes occurs under alkaline condition then reduce as well as oxidized product obtained which is known as disproportional product.
The oxidation of aldehydes occur through potassium dichromate, potassium permanganate or many more. The oxidation of aldehydes in the presence of base is known as cannizzaro's reaction.
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Mole = Mass / Molar mass
6.12 moles = Mass / 74.92 g/mol
Mass = 6.12 moles x 74.92 g/mol
Mass = 458.51g
The compound AB₂ is a covalent compound.
<h3>What is the nature of the molecular substance AB₂?</h3>
The nature of a substance depends on the component elements and the nature of the bonds present in the substance.
Substances can be ionic in nature if ionic bonds are present in the compound or covalent in nature if covalent bonds are present.
The substance AB₂ is composed of the elements A and B where A is a group 14 element and B is a group 16 element.
Group 14 elements are usually non-metals or semi-metals.
Group 16 elements are non-metals.
The bond between A and B will be covalent. Therefore, the compound AB₂ is a covalent compound.
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Answer:
0.0626 M
Explanation:
Equation of the reaction
CH3COOH(aq) + NaOH(aq) ---------> CH3COONa(aq) + H2O
Concentration of acid CA= ???
Concentration of base CB= 0.1943 M
Volume of acid VA= 10.0ml
Volume of base VB= 32.22 ml
Number of moles of acid NA= 1
Number of moles of base NB= 1
From
CA= CB VB NA/VA NB
Hence ;
CA= 0.1943 M × 32.22 ml × 1/10.0ml ×1
CA= 0.0626 M