The question is incomplete, here is the complete question:
Consider the reaction 
Initially, ![[NH_3(g)]=[O_2(g)]](https://tex.z-dn.net/?f=%5BNH_3%28g%29%5D%3D%5BO_2%28g%29%5D)
= 3.60 M; at equilibrium ![[N_2O_4(g)]=0.60M](https://tex.z-dn.net/?f=%5BN_2O_4%28g%29%5D%3D0.60M)
. Calculate the equilibrium concentration for 
<u>Answer:</u> The equilibrium concentration of ammonia is 2.8 M
<u>Explanation:</u>
We are given:
Initial concentration of ![[NH_3(g)]](https://tex.z-dn.net/?f=%5BNH_3%28g%29%5D)
= 3.60 M
Initial concentration of ![[O_2(g)]](https://tex.z-dn.net/?f=%5BO_2%28g%29%5D)
= 3.60 M
For the given chemical equation:
Initial: 3.60 3.60
At eqllm: 3.60-4x 3.60-7x 2x 6x
We are given:
Equilibrium concentration of ![[N_2O_4(g)]](https://tex.z-dn.net/?f=%5BN_2O_4%28g%29%5D)
= 0.60 M
Evaluating the value of 'x'
So, equilibrium concentration of 
Hence, the equilibrium concentration of ammonia is 2.8 M
Answer:
=> 2.8554 g/mL
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 16.59 g
Volume (v) = 5.81 mL
From our question, we are to determine the density (rho) of the rock.
The formula:
Substitute the values into the formula:
= 2.8554 g/mL
Therefore, the density (rho) of the rock is 2.8554 g/mL.
The best description of Ernest Rutherford's experiment is letter C. The positively charged particles were fired through a gold foil.
