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GalinKa [24]
3 years ago
5

Ch3cooh (acetic acid) can form hydrogen bonds between its molecules. based on the lewis structure shown below, how many hydrogen

bond donor and acceptor atoms does this molecule have?

Chemistry
1 answer:
never [62]3 years ago
4 0
Answer:
           Number of Hydrogen Bond Acceptor atoms  =  2

           Number of Hydrogen Bond Donor atoms  =  1

Explanation:
                   Hydrogen bond interactions are formed between those molecules which contains partial positive hydrogen atoms bonded covalently to most electronegative atoms like Oxygen, Nitrogen and Fluorine. 
                   When hydrogen is attached to Oxygen, Nitrogen or Fluorine its electron density decreases and gets partial positive charge, this partial positive charged hydrogen atom then makes hydrogen bonding with the most electronegative element (partial negative) of neighbor molecule.
                    In Acetic acid there are two oxygen atoms hence there are two most electronegative elements therefore, two Hydrogen Bond Acceptor atom and each oxygen atom can accept two hydrogen bonds.
                    Also, it contains only one Hydrogen atom attached to oxygen atom so it has one Hydrogen Bond Donor atom.

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A) CH4

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<h2>Further explanation </h2>

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Details

Class: college

Subject: chemistry

Keywords: ch4, methane, chemicals

8 0
3 years ago
Calculate the density of a material that has a mass of 50.5g and a volume of 14.5cm3.
White raven [17]
Density = mass / volume = 50.5/14.5 = 3.48 g/cm ^{3}.
4 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

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