Luminescent I'm pretty sure :)
Answer:
ΔH°r = -1562 kJ
Explanation:
Let's consider the following combustion.
C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)
We can calculate the standard heat of reaction (ΔH°r) using the following expression:
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(i) are the standard heats of formation of reactants and products
The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]
ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]
ΔH°r = -1562 kJ
Answer:
i think this would be an example of a physical change because here the substance is changing directly from a solid to a gas without passing through the liquid state which means that the composition of the substance has not changed and remains the same. This is also called sublimation.
That is a bond between carbon and chlorine, as in chloro methane:
H3C-cl
i hope i helped you❤
Answer:
63.02 g.
Explanation:
- Na reacts with Cl₂ according to the balanced equation:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear that 2 mole of Na react with 1 mole of Cl₂ to produce 2 moles of NaCl.
- Firstly, we need to calculate the no. of moles of Na and Cl₂:
no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.
no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.
- From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).
<em>So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".</em>
The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.
∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.
- To get the no. of grams of produced NaCl:
<u><em>using cross multiplication:</em></u>
2 mol of Na produce → 2 mol of NaCl, from the stichiometry.
∴ 0.826 mol of Na produce → 0.826 mol of NaCl.
∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.
∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.
