Answer:
"They drink water when it is aivailable but can go weeks without it, they rely on morning dew and the water content of their food". they can use plants as resources to live.
It is kept constant
There is the answer if it helped
<u>Answer:</u>
<u>For a:</u> The chemical equation for the dissolution of sodium carbonate is 
<u>For b:</u> The net acid-base reaction is 
<u>Explanation:</u>
Dissolution reaction is defined as the reaction in which a solid compound gets dissolved in water to form aqueous solution.
The chemical equation for the dissolution of sodium carbonate follows:

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.
The chemical equation for the ionization of sodium carbonate follows:

Now, the anion formed which is
reacts with water to form conjugate acid.
The chemical equation for the reaction of anion with water follows:

Hence, the net acid-base reaction of the anion formed and water is written above.
Answer:
A) 4.3 × 10²⁴ atoms
Explanation:
Step 1: Given data
Moles of neon: 7.2 moles
Step 2: Calculate the number of atoms present in 7.2 moles of neon
In order to convert moles to toms, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ neon atoms in 1 mole of neon atoms.
7.2 mol × 6.02 × 10²³ atoms/mol = 4.3 × 10²⁴ atoms
Answer: (a) The solubility of CuCl in pure water is
.
(b) The solubility of CuCl in 0.1 M NaCl is
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.

Initial: 0 0
Change: +x +x
Equilibm: x x

And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)

x = 
Hence, the solubility of CuCl in pure water is
.
(b) When NaCl is 0.1 M,
, 
, 
Net equation: 
= 0.1044
So for, 
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x = 
Therefore, the solubility of CuCl in 0.1 M NaCl is
.