Which is smallest to biggest glucose,starch,protein,carbon,water,amino,oxygen
1 answer:
Answer:
Smallest to biggest order is as follows.
C < O < Amino < Water < Glucose < Starch < protein.
Explanation:
The given atoms or molecules and their atomic masses is as follows.
Glucose - - 180 g/mol
Starch - - 692.7 g/mol
Protein - - it is vary for different proteins.
Carbon - C - 12 g/mol
Water - - 18 g/mol
Amino - - 17 g/mol
Oxygen - O - 16 g/mol
Increase the molar mass increases size of the molecules.
Therefore, Smallest to biggest order is as follows.
C < O < Amino < Water < Glucose < Starch < protein.
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Answer:
The enthalpy change for the solution process = - 158.34 kJ/mol
Explanation:
Given that:
The mass of salt AX = 2.499 g
The mass of water = 135.3 g
The mass of the solution = ( 2.499 + 135.3 ) g = 137.799 g
The specific heat of salt solution s is known to be = 4.18 J/g° C
The change in temperature i.e. ΔT = 35.2 °C - 23.6 °C = 11.6 °C
Thus, the amount of heat raised is equal to the heat absorbed by the calorimeter.
∴
Recall that the mass of the salt = 2.499 g
The number of moles of the salt =
= 0.0422 mol of AX
Finally the enthalpy change,
= - 158.34 kJ/mol
The enthalpy change for the solution process = - 158.34 kJ/mol
Answer:
A. 0.143 M
B. 0.0523 M
Explanation:
A.
Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).
KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄
The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:
1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol
The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.
5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:
M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M
B.
Let's consider the neutralization of potassium hydroxide and perchloric acid.
KOH + HClO₄ → KClO₄ + H₂O
When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.