Answer:
(2) Boiling point order: Methanol (highest) > Methanal > Methane (weakest)
(3) Boiling point of alcohol will be higher than ester molecules
Explanation:
(2) Methane is a non-polar molecule. Hence only weakest van der waal inter molecular force is present between methane molecules.
Methanal is polar molecule due to presence of polar aldehyde group. hence weaker dipole-dipole inter molecular force is present between methanal molecules.
Methanol is a polar protic molecule. Hence strongest H-bonding force act between methanol molecules.
The stronger the inter molecular force, the higher will be boiling point.
Boiling point order: Methanol (highest) > Methanal > Methane (weakest)
(3) An alcohol is a polar protic molecule. Hence strongest H-bonding force exist between alcohol molecules.
An ester is polar molecule. Therefore weaker dipole-dipole inter molecular force is present between ester molecules.
So boiling point of alcohol will be higher than ester molecules.
Answer:
an
Explanation:
It is AN because investigation starts with I ( i is a vowel). When there are vowels we use An.
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Volume is mass divided by density or mass/density.
so 244/2.70
is equal to 90.3703703704 which is the volume
Answer is: K <span>be for the reaction at 375 K is 326.
</span>Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
</span><span>Δ<span>H = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K</span></span></span>·mol.<span>
K</span>₁ = 6,8·10⁵.<span>
K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.