Answer:
Cu(IO3)2 + K2CO3 → 2KIO3 + CuCO3
2Al₂O₃ = 4Al + 3O₂
M(A₂O₃)=101.96 g/mol
m(Al₂O₃)=250 g
n(O₂)=3m(Al₂O₃)/{2M(Al₂O₃)}
n(O₂)=3*250/{2*101.96}=3.678 mol
Answer:
V₂ = 16.5 L
Explanation:
To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:
V₁/n₁ = V₂/n₂
In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].
- n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol
<em>We use the reaction to calculate n₂</em>:
2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)
0.1900 mol CO *
0.1900 mol CO₂
0.1900 mol NO *
0.095 mol N₂
- n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol
Calculating V₂:
22.0 L / 0.3800 mol = V₂ / 0.2850 mol
V₂ = 16.5 L
Mud is a suspension because it is a heterogeneous mixture that has many solid particles that have sedimented in various parts of the mixture, meaning they can be easily seen or removed. A solution is homogeneous like putting sugar in water or similar.
Answer: There are 0.00269 moles of acetic acid in buffer.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of acetic acid solution = 0.0880 M
Volume of solution = 30.6 mL
Putting values in equation 1, we get:

Thus there are 0.00269 moles of acetic acid in buffer.