n = m / M
Where, n is moles of the compound (mol), m is the mass of the compound (g) and M is the molar mass of the compound (g/mol)
Here, the given ethanol mass = 50.0 kg = 50.0 x 10³ g
Molar mass of the ethanol = (12 x 2 + 1x 6 + 1 x 16) g/mol
= 46 g/mol
Hence, moles in 50.0kg of ethanol = 50.0 x 10³ g / 46 g/mol
= 1086.96 mol
Answer:
eheheehehehszndn!jejxxnndrrjrrrfufurururufjththjrjrjdjjjrj\u\ujrjeejrjjjj carbon
Answer:
ΔHr = -103,4 kcal/mol
Explanation:
<u>Using:</u>
<u>AH° (kcal/mol)
</u>
<u>Metano (CH)
</u>
<u>-17,9
</u>
<u>Cloro (CI)
</u>
<u>tetraclorometano (CCI)
</u>
<u>- 33,3
</u>
<u>Acido cloridrico (HCI)
</u>
<u>-22</u>
It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:
ΔHr = (ΔH products - ΔH reactants)
For the reaction:
CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)
The balanced reaction is:
CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)
The ΔH's of formation for these compounds are:
ΔH CH₄(g): -17,9 kcal/mol
ΔH Cl₂(g): 0 kcal/mol
ΔH CCl₄(g): -33,3 kcal/mol
ΔH HCl(g): -22 kcal/mol
The ΔHr is:
-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)
<em>ΔHr = -103,4 kcal/mol</em>
<em></em>
I hope it helps!
Glucose has empirical formula C6H12O6. So its formula mass can be calculated from that: 12.01x6 + 1.008x12 + 16.00x6 = 72.06 + 12.096 + 96.00 = 180.156 which needs to be rounded to two decimals to get 180.16 g/mole<span>.</span>