1: a, 2:a, 3:a, 4:c, 5:d, 6:d, 7:c;
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
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Answer: 568g/mol
Explanation:
It should be noted that there are 40 atoms of carbon in lycopene.
Since mass of 1 carbon = 12g/mol
Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol
Let the molar mass of lycopene be represented by x.
Therefore the molar mass of carbon = x × mass percent of carbon in lycopene
x × 84.49% = 480g/mol
x × 0.8449 = 480g/mol
x = 480/0.8449
x = 568g/mol
The molar mass of lycopene is 568g/mol
We are told that there are 1.55 x 10²³ molecules of Cl₂ and we need to calculate the mass of these molecules. We need to do several conversions. The easiest will be to convert the amount of molecules to the number of moles present. To do this, we need to use Avogadro's number which is 6.022 x 10²³ molecules/mole.
1.55 x 10²³ molecules / 6.022 x 10²³ molecules/mole = 0.257 moles Cl₂
Now that we have the moles of Cl₂ present, we can convert this value to a mass of Cl₂ by using the molecular mass of Cl₂. The molecular mass is 70.906 g/mol.
0.257 moles Cl₂ x 70.906 g/mol = 18.3 g Cl₂
Therefore, 1.55 x 10²³ molecules of Cl₂ will have a mass of 18.3 g.
