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3 years ago

12

Balance the following equations by inserting the proper coefficients.

2 answers:

kozerog [31]3 years ago

7 0

Answer : The given balanced equations will be,

$CH_4+2O2\rightarrow CO_2+2H_2O$

$CaCl_2+2AgNO_3\rightarrow Ca(NO_3)_2+2AgCl$

$C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O$

Explanation :

Balanced equation : In the balanced equation, the individual elements present on reactant side must be equal to the elements present on the product side.

The given balanced equations will be,

$CH_4+2O2\rightarrow CO_2+2H_2O$

$CaCl_2+2AgNO_3\rightarrow Ca(NO_3)_2+2AgCl$

$C_2H_6O+3O_2\rightarrow 2CO_2+3H_2O$

Montano1993 [528]3 years ago

6 0

1212. 1212. 1323........

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In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

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Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

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$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

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