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mash [69]
4 years ago
10

What type of molecular attraction accounts for the high boiling point of H2O

Chemistry
1 answer:
aleksklad [387]4 years ago
3 0

Hydrogen bonding

Explanation:

Hydrogen bonding between water molecules account for its high boiling point.

  • Intermolecular forces determines the bulk of the physical properties of matter.
  • Hydrogen bond is a special type of dipole dipole attraction
  • It forms when a hydrogen on one molecule attracts an electronegative atom on another usually Oxygen, fluorine and nitrogen.
  • These intermolecular forces are very strong and difficult to break.

Learn more:

Hydrogen bond brainly.com/question/10602513

#learnwithbrainly

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In water is the mass of one molecule equal to the mass of one molecule in steam
inna [77]

Answer:

The mass of one mole of water molecule = 18 ×N  

A  amu

Explanation: The mass of one mole of water molecule = Mass of one molecule ×N A Also mass of one mole of water is its molar mass in grams. So the mass of one mole of water is 18 g

​

3 0
3 years ago
Describe the three ways heat can be transferred-radiation, conduction, and convection
hjlf
An example of radiation can be from a fire since the firs is letting off heat without contact. An example of convection can be boiling water since there are molecules moving in the water (a fluid). An example of conduction can be a tea spoon in a cup of hot tea, the heat is getting transferred froma hot object to a cool object.
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4 years ago
Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
MariettaO [177]

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

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3 years ago
How many atoms of Hydrogen are in 67.2 L of H2 at STP?​
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