Answer:
T₂ = 150 K
Explanation:
Given data:
Initial volume of gas = 804 mL
Initial temperature = 27°C (27+273=300 K)
Final temperature = ?
Final volume = 402 mL
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁/V₁
T₂ = 402 mL × 300 K / 804 mL
T₂ = 120,600 mL.K / 804 mL
T₂ = 150 K
There are two in the first ring leaving only six out of eight to go on the next
Answer:
[H₂SO₄] = 0.255M
Explanation:
H₂SO₄ + 2KOH => K₂SO₄ + 2H₂O
2(M x V)acid = (M x V)base => M(acid) = (M x V)base /2·Vacid
[H₂SO₄] = (0.5M)(25.5ml)/2(25.0ml) = 0.255M
Answer:
-431.5 J/g
Explanation:
Mass of solution = Mass of solute + mass of solvent
Solute is KOH while solvent is water.
Mass of KOH = 16.9 g
Mass of water = 90.8 g
Mass of solution = 16.9 + 90.8
= 107.7 g
Change in temperature (Δt) = 34.27 - 18.5
= 16.2 °C
Heat required to raise the temperature of water is released by dissolving KOH.
Therefore,
Heat released by KOH = m × s× Δt
= 107.7 × 4.18 × 16.2
= 7293 J
Heat released by per g KOH = 7293 J/16.9 g
= 431.5 J/g
As heat is released therefore, enthalpy change would be negative.
Enthalpy change of KOH = -431.5 J/g
