2nd
Explanation:
because i guesses i guess so yeah trust your gut buddy
 
        
             
        
        
        
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
 
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
 
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
 
        
             
        
        
        
Answer:
Mass = 13.23 g  
Explanation:
Given data:
Mass of oxygen = 48.0 g
Mass of propane burn = ?
Solution:
Chemical equation:
C₃H₈ + 5O₂     →      3CO₂ + 4H₂O
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 48.0 g/ 32 g/mol
Number of moles = 1.5 mol
now we will compare the moles of propane and oxygen.
               O₂           :          C₃H₈
                5            :            1
              1.5            :          1/5×1.5 = 0.3 mol
Mass of propane burn:
Mass = number of moles × molar mass
Mass = 0.3 mol × 44.1 g/mol
Mass = 13.23 g