<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
Answer:
454.3 g.
Explanation:
1.0 mol of CaO liberates → – 64.8 kJ.
??? mol of CaO liberates → - 525 kJ.
∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.
<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>
Answer:
1.403x10²⁴ molecules
Explanation:
In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first<u> convert grams to moles</u> using its <em>molar mass</em>:
- 102.5 g ÷ 44 g/mol = 2.330 mol CO₂
Now we <u>convert moles into molecules </u>using <em>Avogadro's number</em>:
- 2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ molecules
<span>Decomposition reaction.</span>
A nanoparticle is larger than an atom. A nanoparticle is usually made from a few hundred atoms. These particles range from 1 nanometers to 100 nanometers. On the other hand an atom ranges from 0.1 nanometers to 105 nanometers. Using the sizes above, one can clearly see and understand that an atom is smaller.
