Answer:
234.35 °C
Explanation:
Given data:
Volume of balloon = 125000 mL
Moles of oxygen = 3 mol
Pressure = 1 atm
Temperature = ?
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Volume of balloon = 125000 mL × 1 L /1000 mL
Volume of balloon = 125 L
Now we will put the values:
Ideal gas constant = R = 0.0821 atm.L/mol.K
PV = nRT
T = PV/nR
T = 1 atm × 125 L/ 0.0821 atm.L/mol.K × 3 mol
T= 125 /0.2463 /K
T = 507.5 K
K to °C
507.5 K - 273.15 = 234.35 °C
Zero is the value of water potential of pure water at atmospheric pressure.
If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).
<h3>What is base dissociation constant? </h3><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 6.2× 10^(-10)
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) /{6.2×10^(-10) }
= 1.6× 10^(-5)
Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).
learn more about base dissociation constant :
brainly.com/question/9234362
#SPJ4
Answer:
Final Temperature = 36.54 ⁰C
Explanation:
Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,
V₁ / T₁ = V₂ / T₂
Data Given;
V₁ = 32 L
T₁ = 10 °C = 283.15 K ∴ K = °C + 273.15
V₂ = 35 L
T₂ = ??
Solving equation for T₂,
T₂ = V₂ × T₁ / V₁
Putting values,
T₂ = (35 L × 283.15 K) ÷ 32 L
T₂ = 309.69 K ∴ ( 36.54 °C )
Result:
As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.
Answer:
option A
an increase in entropy and a decrease in enthalpy
pls mark brainliest
