Answer:
All carbons in the carbon skeleton contain the maximum number of hydrogen atoms
Explanation:
Saturated fats are class of compounds having all the fatty acids containing only single bonds. In other words, carbon skeleton has only single bonds.
Unsaturated compounds undergoes hydrogenation to form saturated fats.
In hydrogenation, hydrogen atoms are added to the carbon attached to double bond. After become saturated, no more hydrogen atoms can be added.
Therefore, it can be said that in saturated fats all carbons in the carbon skeleton contain the maximum number of hydrogen atoms.
Answer:
HF
Explanation:
Hf has hydrogen bonding which is the strongest intermolecular forces. The stronger the IM forces, the higher the boiling point.
Answer:
The smallest constituent of an element is termed as an atom. According to Jaiden, an atom comprises two subatomic particles and three prime parts, however, this is incorrect as an atom comprises three subatomic particles and two main parts. The three subatomic particles are the protons, electrons, and neutrons. All these parts are smaller in comparison to an atom, thus, they are considered as subatomic particles.
To be considered as an atom, an atom is needed only electrons and neutrons, like hydrogen exhibit one electron and one proton. On the other hand, the overall charge of the atom is not affected by the presence of neutrons, it is the subatomic particle that only enhances the mass of the atom.
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
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1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %