What is the correct name for the cyclic alkane shown on the right? 1,3-dimethyl, 3-ethylcyclohexane 1-ethyl, 1,4-dimethylcyclohexane 1-ethyl-1,3-dimethylcyclohexane 1-ethyl-1,3-dimethylcyclopentane 1,3 dimethyl, 3- ethylcyclopentane A
Methane is the compound CH4, and burning it uses the reaction:
CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.
Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.
n = 9.5/16.042 = 0.592195 mol
Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>
a. 34 mL; b. 110 mL
a. A tablet containing 150 Mg(OH)₂
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂
= 2.572 mmol Mg(OH)₂
<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]
= 5.144 mmol HCl
Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl
b. A tablet containing 850 mg CaCO₃
CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O
<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃
= 8.492 mmol CaCO₃
<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]
= 16.98 mmol HCl
Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl
Answer:
44.7 kWh
Explanation:
Let's consider the reduction of Al₂O₃ to Al in the Bayer process.
6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻
We can establish the following relations:
- The molar mass of Al is 26.98 g/mol.
- 2 moles of Al are produced when 6 moles of e⁻ circulate.
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
- 1 V = 1 J/c
- 1 kWh = 3.6 × 10⁶ J
When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:
40.28 sorry if it is wrong:) but I tried
