Answer:
- 178 ºC
Explanation:
The ideal gas law states that :
PV = nRT,
where P is the pressure, V is the volume, n is number of moles , R is the gas constant and T is the absolute temperature.
For the initial conditions :
P₁ V₁ = n₁ R T₁ (1)
and for the final conditions:
P₂V₂= n₂ R T₂ where n₂ = n₁/2 then P₂ V₂ = n₁/2 T₂ (2)
Assuming V₂ = V₁ and dividing (2) by Eqn (1) :
P₂ V₂ = n₁/2 R T₂ / ( n₁ R T₁) then P₂ / P₁ = 1/2 T₂ / T₁
4.10 atm / 25.7 atm = 1/2 T₂ / 298 K ⇒ T₂ = 0.16 x 298 x 2 = 95.1 K
T₂ = 95 - 273 = - 178 º C
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</h3>
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
learn more about base dissociation constant:
brainly.com/question/9234362
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Answer:
True
Explanation:
Thrust is the force which moves an aircraft through the air. Thrust is used to overcome the drag of an airplane, and to overcome the weight of a rocket. Thrust is generated by the engines of the aircraft through some kind of propulsion system.
