<u>Answer:</u> The molar solubility of 
is 
<u>Explanation:</u>
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The balanced equilibrium reaction for the ionization of calcium fluoride follows:
s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
We are given:
Putting values in above equation, we get:
Hence, the molar solubility of is
I believe the answer is: Protons and Neutrons
Protons weigh 1 amu while neutrons also weigh 1 amu. I’m not sure if they are talking about how many there is.
Answer:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring = 8 g/mL
Explanation:
From the question given above, the following data were obtained:
Mass of ring = 32 g
Volume of water = 64 mL
Volume of water + ring = 68 mL
Density of ring =?
Next, we shall determine the volume of the ring. This can be obtained as follow:
Volume of water = 64 mL
Volume of water + ring = 68 mL
Volume of ring =?
Volume of ring= (Volume of water + ring) – (Volume of water)
Volume of ring = 68 – 64
Volume of ring = 4 mL
Finally, we shall determine the density of the ring. This can be obtained as follow:
Mass of ring = 32 g
Volume of ring = 4 mL
Density of ring =?
Density = mass / volume
Density of ring = 32 / 4
Density of ring = 8 g/mL
Um im pretty sure there are only about 100 different atoms...
Answer:
Q = 30355.2 J
Explanation:
Given data:
Mass of ice = 120 g
Initial temperature = -5°C
Final temperature = 115°C
Energy required = ?
Solution:
Specific heat capacity of ice is = 2.108 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 -T1
ΔT = 115 - (-5°C)
ΔT = 120 °C
Q = 120 g × 2.108 j/g.°C × 120 °C
Q = 30355.2 J
