Answer: Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.
Explanation:
Hope this helps!
First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5
The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
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.
Mass of water produced : 0.146 g
<h3>Further explanation</h3>
Given
33.2 mL of 0.245 M lithium hydroxide
Required
mass of water
Solution
Reaction
HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)
mol LiOH :
= M x V
= 0.245 x 33.2 ml
= 8.134 mmol
From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol
mass H₂O :
= mol x MW
= 8.134 x 10⁻³ mol x 18 g/mol
= 0.146 g
