Answer:
If the pKa of the acid is low (negative), then the acid is strong.
Explanation:
Ka, <em>the acid ionization constant, </em>measures the strength of an acid in a solution. Stronger acids have higher Ka values.
We defined: pKa = -log[Ka]
This function is a decreasing function, meaning that pKa will be getting smaller and smaller, while increasing Ka (high values of Ka will have negative pKa values). Therefore, stronger acids (high values of Ka), will have low (negative) pKa values.
A positive acceleration indicates that the object sped up. This means that if you compare the first speed to the second, the second speed should be higher.
A negative acceleration indicates that the object has slowed down. This means that if you compare the first speed to the second, the second speed should be lower.
If an acceleration is 0, it means that it neither slowed down nor sped up.
Now let us analyze your problem by listing down the speed and the time:
At noon: 4 mi/hr
12:30 : 6 mi/hr
2:30 : 2 mi/hr
From noon to 12:30, you will notice that there is an increase in speed. This means that Tommy had a positive acceleration. (Rules out D.)
From 12:30 to 2:30, there is a decrease in speed. This would indicate that Tommy had a negative acceleration. (Rules out C.)
No speed was the same, so acceleration was never 0. (Rules out A.)
From the assumptions above, we can now deduce that the answer is B.
The answer is definitely not 70.923
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B8%5Ctimes%20H_f_%7BCO_2%7D%2B10%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B2%5Ctimes%20H_f_%7BC_4H_%7B10%7D%2B13%5Ctimes%20H_f_%7BO_2%7D%7D%5D)
![\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.7%29%2B%2813%5Ctimes%200%29%5D)
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
