1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(HCl) = 1 : 1.
0,00675 mol : n(HCl) = 1 : 1.
n(HCl) = 0,00675 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0,00675 mol ÷ 0,130 mol/L.
V(HCl) = 0,0519 L = 51,92 ml.
2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(KCl) = 1 : 1.
0,00675 mol : n(KCl) = 1 : 1.
n(KCl) = 0,00675 mol.
m(KCl) = n(KCl) · M(KCl).
m(KCl) = 0,00675 mol · 74,55 g/mol.
m(KCl) = 0,503 g.
n - amount of substance.
M - molar mass.
The pressure of the gas used in the weather balloon increases to expand the balloon.
Explanation:
- Weather balloons contain the boxes where the weather measurement instruments are present that is attached to the large balloon.
- Weather balloon uses gases like Hydrogen or Helium. When the weather balloon rises to the atmosphere, the air pressure decreases. This leads to the increase in the pressure of hydrogen or Helium gas used in the weather balloon. This expands the balloon.
- The gas particles hits the balloon container and generates the pressure. The increase of pressure thus helps the weather balloon to move in a constant speed through the atmosphere.
Answer:
Explanation:
From the information given:
no of moles of 
= 0.01 L × 0.0010 mol/L
no of moles of = 
no of moles of 
= 0.01 L × 0.00010 mol/L
no of moles of = 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2
