Explanation:
Tollens' reagent is prepared by using two-step process : -
Step 1:
Silver oxide is formed by mixing aqueous silver nitrate with base like sodium hydroxide. The reaction is shown below as:

Step 2
Ammonia solution is drop-wise added until all the silver oxide dissolves to form the reagent. The reaction is shown below as:

Answer: The issue of long-term effects of the Chernobyl disaster on civilians is very controversial. The number of people whose lives were affected by the disaster is sizable. Over 300,000 people were resettled because of the disaster; millions lived and continue to live in the contaminated area.
Explanation:
8.98
×
N
A
cobalt atoms
Explanation:
N
A
,
Avogadro's number
specifies
6.0221
×
10
23
individual particles. It is simply another collective number like a dozen, or a score, or a gross.
N
A
has the property that
6.0221
×
10
23
individual cobalt atoms has a mass of
58.93
⋅
g
. How did I know that? Did I have it memorized?
So the quantity is
≈
54
×
10
23
cobalt atoms.
Answer:
11.39
Explanation:
Given that:


Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:

Thus,


Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)


Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)

x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
Answer:
Kc = 2.34 mol*L
Explanation:
The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.
A + B ⇄ C + D
Kc = [C] * [D] / [A] * [B]
According to the reaction
Kc = [SO2]^2 * [O2]^2 / [SO3]^2
Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3
0.450 --> 0 + 0 (Beginning of the reaction)
0.260 --> 0.260 + 0.130 (During the reaction)
0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)
Kc = [0.260]^2 + [0.130]^2 / [0.190]^2
Kc = 2.34 mol*L