Answer:
<em>Your</em><em> </em><em>Answer</em><em> </em><em>is</em><em> </em><em>Option</em><em> </em><em>A</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>Gas</em><em>.</em>
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
Data Given:
Pressure = P = ?
Volume = V = 3.0 L
Temperature = T = 115 °C + 273 = 388 K
Mass = m = 75.0 g
M.mass = M = 44 g/mol
Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M
n = 75.0 g / 44 g.mol⁻¹
n = 1.704 mol
Putting Values in Eq. 1,
P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L
P = 18.08 atm
I think is 6.588579795x10^13 Kg because the equation is E=mc^2 and E is your Joules and c^2= 9x10^16 so m=(c^2)/E
