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soldi70 [24.7K]
3 years ago
10

2h2 o2 = 2h2o how many moles of water are produced from 13.35 mol of oxygen?

Chemistry
1 answer:
HACTEHA [7]3 years ago
5 0
2 H2 + O2 = 2 H2O

1 mole O2 --------------- 2 moles H2O
13.35 moles O2 -------- ( moles H2O)

moles H2O = 13.35 x 2 / 1

= 26.7 moles of H2O

hope this helps!
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What is the mass of an object with a net force of 10 N accelerating at 2 m/s^2??
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Hydrogen bonds are stronger than the dipole dipole attraction force present in any molecule.

<h3>What is bonding in molecules?</h3>

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What is the entropy change of the surroundings
KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

5 0
3 years ago
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