Answer:
V₂ = 2.1 L
Explanation:
Given data:
Initial volume of balloon = 2.0 L
Initial temperature = 25°C
Final temperature = 35°C
Final volume of balloon in hot room = ?
Solution:
Initial temperature = 25°C (25+273= 298 K)
Final temperature = 35°C (35+273 = 308 k)
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
V₂ = V₁T₂/T₁
V₂ = 2.0 L × 308 K / 298 k
V₂ = 616 L.K / 298 K
V₂ = 2.1 L
I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,
2H2 + O2 --> 2H2O
The masses of each of the reactants are calculated below.
2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g
Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2
Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.
<em>Answer: 1.22 g of oxygen</em>
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
