Answer:
Mass of water produced is 22.86 g.
Explanation:
Given data:
Mass of hydrogen = 2.56 g
Mass of oxygen = 20.32 g
Mass of water = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 20.32 g/ 32 g/mol
Number of moles = 0.635 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 2.56 g/ 2 g/mol
Number of moles = 1.28 mol
Now we will compare the moles of water with oxygen and hydrogen.
O₂ : H₂O
1 : 2
0.635 ; 2×0.635 = 1.27
H₂ : H₂O
2 : 2
1.28 : 1.28
The number of moles of water produced by oxygen are less thus it will be limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 1.27 × 18 g/mol
Mass = 22.86 g
Answer:
<h2>Volume = 9.29 mL</h2>
Explanation:
Density of a substance can be found by using the formula

From the question
Density = 11.3 g/mL
mass = 105 g
Substitute the values into the above formula and solve for the volume
That's

Cross multiply
11.3v = 105
Divide both sides by 11.3

v = 9.29203
We have the final answer as
<h3>Volume = 9.29 mL</h3>
Hope this helps you
Answer:
The degree of dissociation of acetic acid is 0.08448.
The pH of the solution is 3.72.
Explanation:
The 
The value of the dissociation constant = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)

Initial concentration of the acetic acid = [HAc] =c = 0.00225
Degree of dissociation = α

Initially
c
At equilibrium ;
(c-cα) cα cα
The expression of dissociation constant is given as:
![K_a=\frac{[H^+][Ac^-]}{[HAc]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BAc%5E-%5D%7D%7B%5BHAc%5D%7D)



Solving for α:
α = 0.08448
The degree of dissociation of acetic acid is 0.08448.
![[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%20%3D%200.00225M%5Ctimes%200.08448%3D0.0001901%20M)
The pH of the solution ;
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![=-\log[0.0001901 M]=3.72](https://tex.z-dn.net/?f=%3D-%5Clog%5B0.0001901%20M%5D%3D3.72)
<span>0.453 moles O2 is the solution</span>
Gas x would be carbon dioxide.
note/ acid + carbonate —> salt + water + carbon dioxide
the white precipitate would be calcium carbonate. CaCo₃
note/ this is a common eqn u need to remember.
X - CO₂ (carbón dioxide)
Y - CaCo₃ (calcium carbonate)
sodium carbonate is a basic salt