Answer:
No, it is not sufficient
Please find the workings below
Explanation:
Using E = hf
Where;
E = energy of a photon (J)
h = Planck's constant (6.626 × 10^-34 J/s)
f = frequency
However, λ = v/f
f = v/λ
Where; λ = wavelength of light = 325nm = 325 × 10^-9m
v = speed of light (3 × 10^8 m/s)
Hence, E = hv/λ
E = 6.626 × 10^-34 × 3 × 10^8 ÷ 325 × 10^-9
E = 19.878 × 10^-26 ÷ 325 × 10^-9
E = 19.878/325 × 10^ (-26+9)
E = 0.061 × 10^-17
E = 6.1 × 10^-19J
Next, we work out the energy required to dissociate 1 mole of N=N. Since the bond energy is 418 kJ/mol.
E = 418 × 10³ ÷ 6.022 × 10^23
E = 69.412 × 10^(3-23)
E = 69.412 × 10^-20
E = 6.9412 × 10^-19J
6.9412 × 10^-19J is required to break one mole of N=N bond.
Based on the workings above, the photon, which has an energy of 6.1 × 10^-19J is not sufficient to break a N=N bond that has an energy of 6.9412 × 10^-19J
The initial mass of sodium hydroxide is 3.3 g (answer C)
<u><em>calculation</em></u>
Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂
moles = mass÷ molar mass
from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol
moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles
Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)
from given equation NaOH : Fe(OH)₂ is 2 :1
therefore the moles of NaOH = 0.041 x 2 = 0.082 moles
Step 3: find mass of NaOH
mass = moles x molar mass
from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol
mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)
Density is defined as the ratio of mass to the volume.
Density =
(1)
Mass of water = 10 grams
Mass of acetone = 10 grams
Density of water = 1 
Density of acetone = 0.7857 
Put the value of density of water and its mass in equation (1)
1
= 
Volume of water = 10 
Put the value of density of acetone and its mass in equation (1)
0.7857
= 
Volume of acetone = 12.72 
Thus, volume of acetone is more than volume of water because the density of acetone is lower.
Answer : The concentration of
and
are
and
respectively.
Solution : Given,
pH = 4.10
pH : pH is defined as the negative logarithm of hydronium ion concentration.
Formula used : ![pH=-log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D)
First we have to calculate the hydronium ion concentration by using pH formula.
![4.10=-log[H_3O^+]](https://tex.z-dn.net/?f=4.10%3D-log%5BH_3O%5E%2B%5D)
![[H_3O^+]=antilog(-4.10)](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3Dantilog%28-4.10%29)
![[H_3O^+]=7.94\times 10^{-5}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D7.94%5Ctimes%2010%5E%7B-5%7D)
Now we have to calculate the pOH.
As we know, 


Now we have to calculate the hydroxide ion concentration.
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![9.9=-log[OH^-]](https://tex.z-dn.net/?f=9.9%3D-log%5BOH%5E-%5D)
![[OH^-]=antilog(-9.9)](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dantilog%28-9.9%29)
![[OH^-]=1.258\times 10^{-10}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.258%5Ctimes%2010%5E%7B-10%7D)
Therefore, the concentration of
and
are
and
respectively.