Alcohol is a pure substance
I believe the balanced chemical equation is:
C6H12O6 (aq) + 6O2(g)
------> 6CO2(g) + 6H2O(l)
First calculate the
moles of CO2 produced:
moles CO2 = 25.5 g
C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)
moles CO2 = 0.8493 mol
Using PV = nRT from
the ideal gas law:
<span>V = nRT / P</span>
V = 0.8493 mol *
0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm
<span>V = 22.28 L</span>
Answer is: (4) emits energy as it moves to a lower energy state.
Atom emits a characteristic set of discrete wavelengths, according to its electronic energy levels.
Emission spectrum of a chemical element is the spectrum of frequencies emitted due to an atom making a transition from a high energy state to a lower energy state.
Each transition has a specific energy difference.
Each element's emission spectrum is unique.
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
H₂S
<h3>Further explanation</h3>
Given
ΔH fusion and ΔH vaporization of different substances
Required
The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes
Solution
We can use the formula :
Q=heat/energy absorbed
n = moles
The heat absorbed : 58.16 kJ
moles = 3.11
so ΔH vaporization :
The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S
(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)
