Basically, every moles of H2O has around 6.022 x 10^23 Molecules
So for 2.00 moles, the amount of molecules would be :
2x 6.022 x 10^23
= 12.044 x 10^23
Answer:
Explanation:okay siejejeikeje
Answer: The energy of a photon with a wavelength of 820 nm is
.
Explanation:
Given : Wavelength = 820 nm
Convert nm into meter as follows.
![1 nm = 10^{-9}\\So, 820 nm = 820 nm \times \frac{10^{-9} nm}{1 nm}\\= 820 \times 10^{-9} m](https://tex.z-dn.net/?f=1%20nm%20%3D%2010%5E%7B-9%7D%5C%5CSo%2C%20820%20nm%20%3D%20820%20nm%20%5Ctimes%20%5Cfrac%7B10%5E%7B-9%7D%20nm%7D%7B1%20nm%7D%5C%5C%3D%20820%20%5Ctimes%2010%5E%7B-9%7D%20m)
The relation between energy and wavelength is as follows.
![E = \frac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where,
E = energy
h = Planck's constant = ![6.63 \times 10^{-34} kg m^{2}/s](https://tex.z-dn.net/?f=6.63%20%5Ctimes%2010%5E%7B-34%7D%20kg%20m%5E%7B2%7D%2Fs)
c = speed of light = ![3.0 \times 10^{8} m/s](https://tex.z-dn.net/?f=3.0%20%5Ctimes%2010%5E%7B8%7D%20m%2Fs)
= wavelength
Substitute the values into above formula as follows.
![E = \frac{hc}{\lambda}\\= \frac{6.63 \times 10^{-34} kg m^{2}/s \times 3.0 \times 10^{8} m/s}{820 \times 10^{-9} m}\\= \frac{1.989 \times 10^{-25}}{820 \times 10^{-9}}\\= 2.42 \times 10^{-19} m](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%5C%5C%3D%20%5Cfrac%7B6.63%20%5Ctimes%2010%5E%7B-34%7D%20kg%20m%5E%7B2%7D%2Fs%20%5Ctimes%203.0%20%5Ctimes%2010%5E%7B8%7D%20m%2Fs%7D%7B820%20%5Ctimes%2010%5E%7B-9%7D%20m%7D%5C%5C%3D%20%5Cfrac%7B1.989%20%5Ctimes%2010%5E%7B-25%7D%7D%7B820%20%5Ctimes%2010%5E%7B-9%7D%7D%5C%5C%3D%202.42%20%5Ctimes%2010%5E%7B-19%7D%20m)
Thus, we can conclude that energy of a photon with a wavelength of 820 nm is
.
Answer: a. 7.31 g
b. 20.4 %
Explanation:
To calculate the moles :
![\text{Moles of} H_2=\frac{1.29g}{2g/mol}=0.645moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20H_2%3D%5Cfrac%7B1.29g%7D%7B2g%2Fmol%7D%3D0.645moles)
![\text{Moles of} N_2=\frac{9.55}{28g/mol}=0.341moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20N_2%3D%5Cfrac%7B9.55%7D%7B28g%2Fmol%7D%3D0.341moles)
According to stoichiometry :
3 moles of
require = 1 mole of ![N_2](https://tex.z-dn.net/?f=N_2)
Thus 0.645 moles of
will require=
of ![N_2](https://tex.z-dn.net/?f=N_2)
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 3 moles of
give = 2 moles of ![NH_3](https://tex.z-dn.net/?f=NH_3)
Thus 0.645 moles of
give =
of ![NH_3](https://tex.z-dn.net/?f=NH_3)
Mass of ![NH_3=moles\times {\text {Molar mass}}=0.430moles\times 17g/mol=7.31g](https://tex.z-dn.net/?f=NH_3%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.430moles%5Ctimes%2017g%2Fmol%3D7.31g)
Thus theoretical yield for this reaction under the given conditions is 7.31 g.
b) ![{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bpercentage%20yield%7D%7D%3D%5Cfrac%7B%5Ctext%20%7BExperimental%20yield%7D%7D%7B%5Ctext%20%7BTheoretical%20yield%7D%7D%5Ctimes%20100%5C%25)
![{\text {percentage yield}}=\frac{1.49g}{7.31g}\times 100\%=20.4\%](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bpercentage%20yield%7D%7D%3D%5Cfrac%7B1.49g%7D%7B7.31g%7D%5Ctimes%20100%5C%25%3D20.4%5C%25)
The percent yield for this reaction under the given conditions is 20.4 %