Answer:
I not understand your questions
Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
![(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)](https://tex.z-dn.net/?f=%281%29%3AC_6H_6%28l%29%28322K%29%5Crightarrow%20C_6H_6%28l%29%28279K%29%5C%5C%5C%5C%282%29%3AC_6H_6%28l%29%28279K%29%5Crightarrow%20C_6H_6%28s%29%28279K%29%5C%5C%5C%5C%283%29%3AC_6H_6%28s%29%28279K%29%5Crightarrow%20C_6H_6%28s%29%28205K%29)
The expression used will be:
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%5D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene = ![1.51J/g^oC=1.51J/g.K](https://tex.z-dn.net/?f=1.51J%2Fg%5EoC%3D1.51J%2Fg.K)
= specific heat of liquid benzene = ![1.73J/g^oC=1.73J/g.K](https://tex.z-dn.net/?f=1.73J%2Fg%5EoC%3D1.73J%2Fg.K)
= enthalpy change for fusion = ![-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g](https://tex.z-dn.net/?f=-9.8kJ%2Fmol%3D-%5Cfrac%7B9.8%5Ctimes%201000J%2Fmol%7D%7B78g%2Fmol%7D%3D-125.6J%2Fg)
Now put all the given values in the above expression, we get:
![Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]](https://tex.z-dn.net/?f=Q%3D%5B94.4g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28279-322%29K%5D%2B%5B94.4g%5Ctimes%20-125.6J%2Fg%5D%2B%5B94.4g%5Ctimes%201.51J%2Fg.K%5Ctimes%20%28205-279%29K%5D)
![Q=-29427.312J=-29.4kJ](https://tex.z-dn.net/?f=Q%3D-29427.312J%3D-29.4kJ)
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
Answer:
1. 1.72 moles of potassium.
2. All of the answers are true
3. Are used up during a reaction
Explanation:
Recall that the number of moles is obtained from;
Number of moles= Mass of potassium/ molar mass of potassium
Mass of potassium= 67 g
Molar mass of potassium= 39 gmol-1
Number of moles of K= 67 g/ 39 gmol-1
Number of moles = 1.72 moles of potassium.
2. When we look at all the options, we will realize that all the options are true. The rate of reaction doubles for each 10°C rise in temperature, increasing reactant concentration increases particle collision and ultimately increases the rate of reaction. Rate of reaction deals with rate of disappearance of reactants or rate of appearance of products.
3. Catalysts remain unchanged in a chemical reaction because they do not actually participate in the reaction. Hence they are not used up in any chemical reaction.
Answer:
Weight = 98 kg.m/s^2
= 98 N.
Explanation:
Weight of an object is defined as the product of mass of the object and the acceleration due to gravity acting on the object. Its unit is Newton which is denoted as N and is also equivalent to kg.m/s^2.
Mass of the object = 10 kg
Acceleration due to gravity, g = 9.8 m/s^2
Weight = m*g
= 10 kg * 9.8 m/s^2
= 98 kg.m/s^2
= 98 N
20 atoms are in NH4C2H3O2