Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T = 
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where, 
= concentration =
Hence, putting the given values into the above equation as follows.
= 
= 
= 
m
or, = 
= 1 nm (approx)
Also, it is known that = 
Hence, we can conclude that addition of 0.1 of KCl in 0.1 of NaBr "" will decrease but not significantly.
The enthalpy change for an exothermic reaction is negative because heat is being released, so that takes out two of the responses. Since energy is being released into the surroundings due to the exothermic reaction, the potential energy of the products is lower than that of the reactants. Energy is being put in to make the reaction occur, but then that energy is all being released into the surroundings thus a lower potential energy level for the products
Answer:
<h2>The answer is 1.45 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass of seawater = 250 g
volume = 173 mL
It's density is
We have the final answer as
<h3>1.45 g/mL</h3>
Hope this helps you
Answer:
D0wnload Phot0Math........................
Explanation:
Answer: <span>A reaction progress curve has three peaks and two valleys between the peaks. This curve describes a reaction mechanism that involves
<u>three elementary reactions</u>.
Explanation: I have drawn the progress curve with three peaks and two valleys. In fact the peaks shows higher energy and valleys show lower energies. So, Let suppose we react
A and
B. This reaction between A and B results in the formation of
C. In this reaction the energies of A and B are less, and during the progress of reaction they cross a transition state of higher energy and forms product C with lower energy which is present at lower valley. This was first reaction. Other two reactions will be followed by conversion of C to
D and conversion of D into
E.</span>
