O2 is the limited reactant
This is an incomplete question, here is a complete question.
Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.
CaCO₃, Ksp = 8.7 × 10⁻⁹
Answer : The solubility of CaCO₃ is, 
Explanation :
As we know that CaCO₃ dissociates to give 
ion and 
ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ca^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
Let solubility of CaCO₃ be, 's'
Therefore, the solubility of CaCO₃ is,
These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.
Answer:
gvapor or gas to liquid
Explanation:
water which collects as droplets on cold surface when humid air is in contact with it
