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Vinvika [58]
3 years ago
9

C3H8, + 5O2

Chemistry
1 answer:
Anestetic [448]3 years ago
4 0

Answer:

from \: the \: equation \\ 1 \: moles \: of \: propane \: produce \: 4 \: moles \: of \: water \\ 2.50 \:moles \: of \: propane \: will \: produce \: ( \frac{(2.50 \times 4)}{1} ) \: moles \\  = 10 \: moles \: of \: water

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What is the valence electron of manganese?​
DIA [1.3K]

Answer:

25th electron

Explanation:

the last electron is the valence electron. Assuming it has equal numbers of protons and electrons, then the 25th electron is the valence.

6 0
3 years ago
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Can you help me with 5.025- 5.030?
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3 years ago
The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the chan
Sloan [31]

Answer:

The  change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate  change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The  change in entropy in the system is -231.5 J/K

7 0
3 years ago
Water can be heated above its boiling point without boiling. what is this process called?
Hoochie [10]
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5 0
4 years ago
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