We are given the base dissociation constant, Kb, for Pyridine (C5H5N) which is 1.4x10^-9. The acid dissociation constant, Ka for the Pyridium ion or the conjugate acid of Pyridine is to be determined. We know from our chemistry classes that:
Kw = Kb * Ka
where Kw is always equal to 1x10^-14
so, to solve for Ka of Pyridium ion, substitute Kb to the equation together with Kw and solve for Ka:
1x10^-14 = 1.4x10^-9 * Ka
solve for Ka
Ka = 7.14x10^-6
Therefore, the acid dissociation constant of Pyridinium ion is 7.14x10^-6.
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Answer: The correct option is B.
Explanation: To describe the motion of an object, we use the equations of motion.



From the above equations, we require position, speed and direction through which we an calculate the displacement, velocity and acceleration.
To calculate the complete motion of an object, we require all the three factors.
Hence, the correct option is B.
Answer:- 3.
and 
Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.
For example, the molecular formula of benzene is
. The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.
In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is
. In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.
In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is
. The C to H ratio for second molecule is 1:4, so the empirical formula is
. Here also, the empirical formulas are not same and hence it is also not the right choice.
In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is
. In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also
. Hence. this is the correct choice.
In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is
. As the empirical formulas are different, it is not the right choice.
So, the only and only correct pair is the third one. 3.
and 
Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.
Solution : Given,
For Accelerator 1 model,
Input energy = 2078.3 J
Wasted energy = 663.1 J
Output energy = 1415.2 J
For Accelerator 2 model,
Input energy = 7690.0 J
Wasted energy = 2337.5 J
Output energy = 5353.5 J
For Accelerator 3 model,
Input energy = 4061.9 J
Wasted energy = 2259.6 J
Output energy = 1802.3 J
Formula used for lowest percentage of energy lost as waste is:
% energy lost as waste = (Total energy wasted / Total input energy ) × 100
For Accelerator 1 model,
% energy lost as waste =
= 31.90%
For Accelerator 2 model,
% energy lost as waste =
= 30.39%
For Accelerator 3 model,
% energy lost as waste =
= 55.62%
So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.
Answer:
pH= 0.92
Explanation:
HNO3-> H^+ +NO3^-
HNO3 is a strong acid, so it fully dissociates
[HNO3] = 0.12M [H^+] = 0.12M
pH= -log[H^+]
pH=-log[.12] = 0.92
pH = 0.92