Answer:
The correct answer would be - 2.4KJ or, 2400J
Explanation:
Given:
heat capacity of liquid water - 4.18 J/g·°C
heat of vaporization - 40.7 kJ/mol
Mass of water = 1g
Moles of water = mass/molar mass
= 1g/18.016g
= 0.055 moles
Then,
Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)
= m *s*dt + moles * heat of vaporization
= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ
= 137.94J + 2.26KJ
=0.138KJ + 2.26KJ
=2.4KJ or, 2400J
Thus, the correct answer would be - 2.4KJ or, 2400J
Answer:
The correct answer is B the tertiary halides reacts faster than primary halides.
Explanation:
During SN2 reaction the nucleophile attack the alkyl halide from the opposite side resulting in the formation of transition state in which a bond is not completely broken or a new bond is not completely formed.
After a certain period of time the nucleophile attach with the substrate by substituting the existing nuclophile.
An increase in the bulkiness in the alkyl halide the SN2 reaction rate of that alkyl halide decreases.This phenomenon is called steric hindrance.
So from that point of view the that statement tertiary halides reacts faster that secondary halide is not correct.
Answer:
30 °C
Explanation:
When you pour water from one glass to the other, the heat energy will flow from the hot water to the cold.
You have equal masses of water in each glass. The hot water should cool down as much as the cold water warms up.
The final temperature should be halfway between the starting temperatures in each glass.
(10 °C + 50°C)/2 = 60 °C/2 = 30°C
The temperature of the full glass of water should be 30 °C.
Answer : The enthalpy change for this reaction is -193.8 kJ.
Solution :
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{SO_3}\times \Delta H_{SO_3})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{SO_2}\times \Delta H_{SO_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BSO_3%7D%5Ctimes%20%5CDelta%20H_%7BSO_3%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_%7BSO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![\Delta H=[(2\times -395.7)]-[(1\times 0)+(2\times -298.8)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%282%5Ctimes%20-395.7%29%5D-%5B%281%5Ctimes%200%29%2B%282%5Ctimes%20-298.8%29%5D)
Therefore, the enthalpy change for this reaction is, -193.8 KJ
