Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Adding more powdered reactants
Temperature change, colour change, releasing gas, bubbles and change in odor
Answer:
V₂ = 2.96 L
Explanation:
Given data:
Initial volume = 2.00 L
Initial temperature = 250°C
Final volume = ?
Final temperature = 500°C
Solution:
First of all we will convert the temperature into kelvin.
250+273 = 523 k
500+273= 773 k
According to Charles's law,
V∝ T
V = KT
V₁/T₁ = V₂/T₂
V₂ = T₂V₁/T₁
V₂ = 2 L × 773 K / 523 k
V₂ = 1546 L.K / 523 k
V₂ = 2.96 L