Answer:
<span>If a mixture looks smooth and the same throughout it is probably <u>Homogeneous</u>.
Explanation:
Mixture is the combination of different compounds which are unreactive to each other.
Mixture are classified as ...
Solutions; in which the mixed compounds are thoroughly mixed and cannot be distinguished from each other and are said to be homogeneous. In solutions the size of solute is very small (i.e. Less than 1 nm).
Colloids; in which the solute is homogeneous visually but heterogeneous microscopically. The size of particles in this case is between 1 nm to 1 </span>μm.
Suspensions; in which the mixture is heterogeneous, the particle size is greater than 1 μm and settles down (precipitation) under the influence of gravity.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is 
Explanation:
The chemical equation for this decomposition of ammonia is
↔ 
The initial concentration of ammonia is mathematically represented a
![[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%20%20%5Cfrac%7Bn_1%7D%7BV_1%7D%20%20%3D%20%5Cfrac%7B29%7D%7B75%7D)
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
The initial concentration of nitrogen gas is mathematically represented a
![[N_2] = \frac{n_2}{V_2}](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%20%5Cfrac%7Bn_2%7D%7BV_2%7D)
![[N_2] = 0.173 \ M](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%200.173%20%20%5C%20%20M)
So looking at the equation
Initially (Before reaction)


During reaction(this is gotten from the reaction equation )
(this implies that it losses two moles of concentration )
(this implies that it gains 1 moles)
(this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium


Now since
![[NH_3] = 0.387 \ M](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%200.387%20%20%5C%20%20M)
Now the equilibrium constant is
![K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=K_c%20%20%3D%20%20%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
substituting values


Red is the answer in the blank
Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol