Copper (II) Carbonate + Heat yields copper (II) oxide and carbon dioxide
Molecular Equation: CuCo3 + heat > CuO + CO2
First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced.
Mg --> Mg+2
<span>Hg+2 --> Hg+1
</span>
Then you have to balance each half reaction first with electrons before adding them together in one equation

⇒

and

⇒
and then combine them together to form

⇒

It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: A<span>dd H+ and H2O to balance the H's and O's in acidic solution if needed.</span>
Answer:
c. F1-
Explanation:
In this chemical reaction the expression is:
HF + NaF → NaHF2
The ion that always keep the negative charge is the fluorine with a -1, if in this mixture there is more positive ions (H1+) the negative ion (F1-) will join with them.
Remember that also the Cl1- will be free, but the fluorine is more reactive than the fluorine.
Answer:
Reactants are transformed into products
Explanation:
Chemical reactions are happening everywhere, it may be represented by the following equation:
A(aq) + B(s) ⇒ C(g) + D(l)
The left side represent the reactants that interact to get the products.
In the parenthetycal note is the state of the compound, in this case, aquose solution (aq), solid (s), gas (g), liquid (l)
Answer:
pH = 2.18
Explanation:
Perchloric acid (HClO₄) is a strong acid. This means that in an aqueous solution it completely dissociates into H⁺ and ClO₄⁻ species.
First we <u>convert 249 mg HClO₄ into moles</u>, using its <em>molecular weight</em>:
- 249 mg HClO₄ ÷ 100.46 mg/mmol = 2.49 mmol HClO₄
<em>Because it is a strong acid</em>, 2.49 mmol HClO₄ = 2.49 mmol H⁺
We <u>calculate the molar concentration of H⁺</u>:
- 2.49 mmol H⁺ / 380 mL = 6.52x10⁻³ M
Finally we <u>calculate the pH of the solution</u>:
- pH = -log[H⁺] = -log(6.52x10⁻³)