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ira [324]
3 years ago
5

The dipole moment (μ) of hbr (a polar covalent molecule) is 0.797d (debye), and its percent ionic character is 11.8 % . estimate

the bond length of the h−br bond in picometers.note that1 d=3.34×10−30 c⋅m andin a bond with 100% ionic character, q=1.6×10−19
c.
Chemistry
1 answer:
miv72 [106K]3 years ago
6 0
The dipole moment u can be solve using the formula
U = rq
Where u is the dipole moment
R is the bond length
Q = 1.6x10-19 C  
So R = u/q
R = (0.797 d) ( 3.34x10^-30 Cm/ 1 d) /( 1.6x10^-19 C)(0.118)
R = 1.41x10^-10 m
<span>R = 141 pm</span>
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rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}

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(a) From above expression, the rate of consumption of O_2 related to rate of consumption of NH_3 as:

-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}

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Multiply both sides by 6

\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}

So, the rate of formation of H_2O is \frac{6}{4} times that is 1.5 times to the rate of consumption of ammonia.

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