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4 years ago

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Determine the formal charges on the highlighted atoms. if an atom is formally neutral, indicate a charge of zero.the molecule h

3 c c n o. the second c is connected to the c h 3 group with a single bond and to n with a triple bond. n is highlighted and is connected to c with a triple bond and o with a single bond. o is highlighted and is connected to n with a single bond and has 3 lone pairs of electrons.

1 answer:

Deffense [45]4 years ago

7 0

Answer:

Formal charge on n is +1

Formal charge on o is -1

Explanation:

Consider the atom n,

n is connected to o with a single bond and to c with a triple bond

<h3>Formula for formal charge on an atom is number of valence electrons - number of non-bonding electrons - (number of bonding electrons) ÷ 2</h3>

In case of n, number of valence electrons are 5

Number of non-bonding electrons are 0

Number of bonding electrons are 8 (∵ n is attached to 4 bonds and each bond is equivalent to 2 electrons)

∴ Formal charge on n = 5 - 0 - (8 ÷ 2) = 5 - 4 = +1

∴ Formal charge on n is +1

Consider the atom o,

o is connected to n by a single bond

In case of o, number of valence electrons are 6

Number of non-bonding electrons are 6 (∵ it has 3 lone pair of electrons)

Number of bonding electrons are 2 (∵ it is connected to n by a single bond and that bond is equivalent to 2 electrons)

Formal charge on atom o = 6 - 6 - (2 ÷ 2) = -1

∴ Formal charge on o is -1

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3 years ago

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yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

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

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

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Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

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Read 2 more answers

what is the concentration in ppm of a solution which is prepared by dissolving in 15mg of nacl in 200ml water

bezimeni [28]

Answer:

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3 years ago

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pav-90 [236]

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