balance by putting 2 next to HF
Answer:
i am sorry i can not help with that there isnt enough information to work with if you can tell me what you think about the topic i can help you put together a doc
Explanation:
Just find the energy of the <span>blueviolet light with a wavelength of 434.0 nm using the formula:
E = hc / lambda
E = energy
c= speed of light = 3 x 10^8 m/s
h = planck's constant = 6.6 x 10^{-34} m^2 kg / s
lambda = 434 nm = 434 x 10^{-9} m
Putting these values (with appropriate units) in the above formula :
we get: Energy, E = 4.5 x 10^{-19} J
E = 0.45 x 10^{-18} J
Now, the </span>minimum energy is 2.18×10^-{18} J but our energy is 0.45 x 10^{-18} J which is less.
<span>Means the electron will not be removed
</span>
First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
