Answer:
V = 43.95 L
Explanation:
Given data:
Mass of CH₄ decomposed = 15.63 g
Volume of H₂O produced at STP = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → 2H₂O + CO₂
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 15.63 g/ 16 g/mol
Number of moles = 0.98 mol
Now we will compare the moles of H₂O with CH₄.
CH₄ : H₂O
1 : 2
0.98 : 2×0.98 = 1.96 mol
Volume of hydrogen:
PV = nRT
1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K
V = 43.95atm.L / 1atm
V = 43.95 L
The slight positive charges on the hydrogen atoms in water molecules attract the slight negative charges on the oxygen atoms of the other water molecules
Heat. When heat is appplied to liquid it turns into gas.
homeostatic imbalance is the answer, because it's when the internal environment cannot remain in equilibrium.
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
