A submarine fills a special storage area known as a ballast with water in order to increase its density. This allows it to dive
deeper in the ocean. When a submarine needs to rise rapidly, compressed air is used to push the water out of the ballast. What effect does replacing the water in the ballast with air have on the submarine?
2 answers:
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<u>Answer:</u>
<u>For a:</u> The edge length of the crystal is 533.5 pm
<u>For b:</u> The atomic radius of potassium is 231.01 pm
<u>Explanation:</u>
- <u>For a:</u>
To calculate the lattice parameter or edge length of the crystal, we use the equation:
where,
= density =
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal = 39.09 g/mol
= Avogadro's number =
a = edge length of unit cell = ?
Putting values in above equation, we get:
<u>Conversion factor:</u>
Hence, the edge length of the crystal is 533.5 pm
- <u>For b:</u>
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:
where,
R = radius of the lattice = ?
a = edge length = 533.5 pm
Putting values in above equation, we get:
Hence, the atomic radius of potassium is 231.01 pm
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka=
Ka=
Ka=
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= =
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶