Answer:
(c) only Ca2+(aq) and Hg2+(aq)
Explanation:
- In the first step, hydrochloric acid (HCl) is added to the solution. In this case the equilibrium that could take place is:
Ag⁺(aq) + Cl⁻(aq) ↔ AgCl(s)
But no precipitate was formed, so Ag⁺(aq) is absent.
- By adding H₂SO₄(aq) the next equilibrium that could take place is:
Ca⁺²(aq) + SO₄⁻²(aq) ↔ CaSO₄(s)
A white precipitate was formed, so Ca⁺² is present in the solution.
- The following could take place after adding H₂S(aq):
Hg²⁺(aq) + S⁻² ↔ HgS(s)
A black precipitate formed, so Hg⁺² is present as well.
Answer:
The process of dissolving can be endothermic (temperature goes down) or exothermic (temperature goes up).
When water dissolves a substance, the water molecules attract and “bond” to the particles (molecules or ions) of the substance causing the particles to separate from each other.
The “bond” that a water molecule makes is not a covalent or ionic bond. It is a strong attraction caused by water’s polarity.
It takes energy to break the bonds between the molecules or ions of the solute.
Energy is released when water molecules bond to the solute molecules or ions.
If it takes more energy to separate the particles of the solute than is released when the water molecules bond to the particles, then the temperature goes down (endothermic).
If it takes less energy to separate the particles of the solute than is released when the water molecules bond to the particles, then the temperature goes up (exothermic).
Explanation:
Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
13.) would be cytoskeleton and number 16.) would be lysosomes
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka= ![\frac{ [H^{+}] [A^{-} ]}{ [HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BH%5E%7B%2B%7D%5D%20%20%5BA%5E%7B-%7D%20%5D%7D%7B%20%5BHA%5D%7D)
Ka= 
Ka= 
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= 
= 
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
