all of the above is the answer :)
Answer:
Approximately 22.37 days, will it take for the water to be safe to drink.
Explanation:
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
k is rate constant
Given that:- k = 0.27 (day)⁻¹
= 0.63 mg/L
![[A_t]=1.5\times 10^{-3}](https://tex.z-dn.net/?f=%5BA_t%5D%3D1.5%5Ctimes%2010%5E%7B-3%7D)
mg/L
Applying in the above equation as:-
<u>Approximately 22.37 days, will it take for the water to be safe to drink.</u>
Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g) + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
2 NH₃(g) → N₂(g) + 3H₂(g)
Initally 4moles - -
React 2moles 2m + 3m
Eq 2 moles 2m 3m
We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)
The expression for K is: ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
Empirical formula is the simplest ratio of whole numbers of components in a compound
in 100 g of compound
C H O
mass 25.5 g 6.40 g 68.1 g
number of moles 25.5 g/12 g/mol 6.40 g/ 1 g/mol 68.1 g/ 16 g/mol
= 2.13 mol = 6.40 mol = 4.26 mol
divide by least number of moles
2.13/2.13 = 1 6.40/2.13 = 3.0 4.26/2.13 = 2.0
all rounded off
C - 1
H - 3
O - 2
empirical formula - CH₃O₂
