Answer:
a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)
b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)
Explanation:
Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.
a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.
N₂O₅ + H₂O ⇒ HNO₃
Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.
N₂O₅ + H₂O ⇒ 2 HNO₃
b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.
Na₂O + H₂O ⇒ NaOH
Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.
Na₂O + H₂O ⇒ 2 NaOH
Answer:
32.7 g of Zn
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Zn + 2HCl —> ZnCl₂ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂
Next, we shall determine the number of mole of Zn required to produce 0.5 mole of H₂. This can be obtained as follow:
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore, 0.5 mole of Zn will also react to produce to 0.5 mole of H₂.
Thus, 0.5 mole of Zn is required.
Finally, we shall determine the mass of 0.5 mole of Zn. This can be obtained as follow:
Mole of Zn = 0.5 mole
Molar mass of Zn = 65.4 g/mol
Mass of Zn =?
Mass = mole × molar mass
Mass of Zn = 0.5 × 65.4
Mass of Zn = 32.7 g
Thus, 32.7 g of Zn is required to produce 0.5 mole of H₂.
Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)

