Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.
Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.
For example, 
The conjugate base is 
which is a weak base.
Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.
Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
<span>H2PCH3 + H2O <-----> H3PCH3+ + OH-</span>
Cu ions plus EDTA2- ->cu(EDTA)2- plus 2H-
number of moles of CU ions used which is equal to molarity multiplied by volume in litres
that is 50xo.o2 divided by 1000 that is 0.001moles
Since ratio is 1:1 the moles of EDTA is also0.001moles
volume of EDTA is 0.001 divided by 0.0510m which is 0.0196 litres in ml is 0.0196x1000 which is 19.61ml
Answer:
Higher frequency
Explanation:
We can imagine a chemical bond between two atoms as if it were two balls connected by a spring.
According to Hooke's Law, the stretching frequency f is
where µ is the reduced mass of the system
The strength of the bond is analogous to k, the force constant of the spring. Then,
Thus, the stronger the bond, the greater the frequency of vibration.
HCl and NaOH react in a 1:1 ratio, meaning that 1 H+ from HCl will react with 1 OH- from NaOH. Knowing this, and that molarity is mol/liter, all we need to do is use what we have available. First we must find the mols of HCl in our solution, so we set up the following equation in the following steps:
1. 24.75mL x (0.359mol NaOH / 1000mL) = 8.885 x 10^-3mol NaOH
This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
Here we just used the mols of NaOH we found to convert to mols of HCl using the 1:1 ratio described earlier.
From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
8.885 x 10^-3mol HCl/0.03475L = 2.557 x 10^-1M HCl
However, this is the molarity of the HCl and NaOH solution, not the original HCl solution. Using the dilution equation M1V1=M2V2, we can solve for the original molarity.
M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
V1 = the total volume that our mixture has (34.75mL = 0.03475L)
M2 = what we're trying to find
V2 = the amount of the original HCl that we had (10mL = 0.010L)
Simply solving for M2 gives us:
M2 = (M1V1) / V2 or:
M2=((2.557 x 10^-1) x 0.03475L) / 0.010L = 8.89 x 10^-1M HCl. That is your answer.
