Answer:
ΔH = -976.5 kJ
Explanation:
For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.
The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).
So, 24.3 g of benzene has :
n = mass/ molar mass
n = 24.3/78.11
n = 0.311 moles
2 moles ------------ -6278 kJ
0.311 moles ----------- x
By a simple direct three rule:
2x = -1953.08
x = -976.5 kJ
Answer:
Which wind blows cool air inland during the day? Sea Breeze
Which wind blows cool air toward the sea at night? Land Breeze
Which winds blow steadily from specific directions and over long distances? Global Winds
Answer:
Because milk has higher KE than ice, KE is transferred from the milk to the molecules of ice.
Explanation:
The best statement that expresses the transfer of kinetic energy(K.E) is that kinetic energy is transferred from the milk to the ice.
Kinetic energy is form of energy due to motion of the particles of a medium. In this regard, we are dealing with heat energy.
- Heat energy is dissipated from a body at higher temperature to one at a lower temperature.
- Ice is at a lower temperature which is 0°C
- Heat will be transferred in form of thermal energy from the body at higher temperature to one with a lower temperature.
- This is from the milk to the molecules of ice.
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
