Answer:
The true statements are given below.
Explanation:
1 D glucose is a reducing sugar
2 The oxidation of reducing sugar forms a carboxylic acid sugar.
D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.
The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).
Answer:
A - Is the material malleable or ductile?
Explanation:
Answer:
k = 1.3 x 10⁻³ s⁻¹
Explanation:
For a first order reaction the integrated rate law is
Ln [A]t/[A]₀ = - kt
where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.
We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus
Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt
- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 84.4 %
Percentage of H = 15.6 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 84.4 g
Mass of H = 15.6 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.
For Carbon = 
For Hydrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 2
Hence, the empirical formula for the given compound is 
Answer:
Molecular formula = C₆H₁₂O₆
Explanation:
Given data:
Mass of hydrogen = 31.7 g
Mass of carbon = 283.4 g
Mass of oxygen = 377.4 g
Molar mass of compound = 176.124 g/mol
Molecular formula = ?
Solution:
Number of gram atoms of H = 31.7 / 1.01 = 31.4
Number of gram atoms of O = 377.4 / 16 = 23.6
Number of gram atoms of C = 283.4 / 12 = 23.6
Atomic ratio:
C : H : O
23.6/23.6 : 31.4/23.6 : 23.6/23.6
1 : 1.33 : 1
C : H : O =3 (1 : 1.33 : 1
)
Empirical formula is C₃H₄O₃.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+4+3×16 = 88
n = 176.124 / 88
n = 12
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₃H₄O₃)
Molecular formula = C₆H₁₂O₆
