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Phoenix [80]
3 years ago
8

Based on lewis structure analysis, what is the formal charge of the central atom in clf3

Chemistry
1 answer:
MArishka [77]3 years ago
4 0
Formal Charge is calculated as, Lewis structure is attached below,

Formal Charge  =  [# of valence electrons] - [electrons in lone pairs + 1/2 the                                     number of bonding electrons]

# of valence electrons of Cl  = 7

electrons in lone pairs  =  4

number of bonding electrons  =  6

Formal Charge  =  [7] - [4 + 6/2]

Formal Charge  =  [7] - [4 + 3]

Formal Charge  =  [7] - [7]

Formal Charge  =  0

Result:
           Formal charge on Cl in ClF₃ is zero.

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If an object has a density of 0.55 g/mL, what is its density in cg/L?
otez555 [7]

if              1 g is equal to 100 cg

then  0.55 g are equal to X cg

X = (0.55 × 100 ) / 1 = 55 cg

The density of the object is 55 cg/L.

7 0
3 years ago
Read 2 more answers
A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
deff fn [24]

Answer:

A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

Explanation:

From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

            A = Pre exponetial factor

            E_{a} = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

T_{1}=505K

T_{2}=525K

R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
2 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
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How many electrons would an uncharged atom of nitrogen have
dybincka [34]
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3 0
2 years ago
Please help !!!!!!!!
Salsk061 [2.6K]
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2) a) CH4 + Br2 -> CH3Br + HBr
2) b) methane + bromine is substitution because one hydrogen atom from methane is replaced by one bromine atom. addition reaction takes place when one molecule combines with another to form a larger molecule so therefore a molecule from X and bromine combine to form XBr.
6 0
2 years ago
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