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3 years ago

A bottle with 18.02 grams of water contains

Chemistry

2 answers:

8_murik_8 [283]3 years ago

8 0

What are the options?

Romashka [77]3 years ago

7 0

Answer:

Explanation:

nH2O = mass/m.wt

= 18.08/18.08= 1 mole

1 mole contains 6.02x10 23 Avogadro's num of H2O molecules

each molecule has

2mol of hydrogen atom =2 x 6.03x10 23 atoms

1 mole of oxygen has = 1x 6.02x10 23 atoms

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3 years ago

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3 years ago

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Why would a C-O bond have only provide 358 kJ/mol each while C=O provides 799 kJ/mol? Explain in detail.

Alina [70]

Consider the amount of energy needed to brake the bonds.

In c=o there are both a pi-bond and a sigma bond. Therefore it requires a higher energy to brake it. When considering the c-o bond there is only a single sigma bond is needed to be broken. Therefore braking is easier.

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3 years ago

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose

Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624 = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

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3 years ago