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guajiro [1.7K]
3 years ago
10

The permanent electric dipole moment of the water molecule 1H2O2 is 6.2 * 10-30 C m. What is the maximum possible torque on a wa

ter molecule in a 5.0 * 108 N/C electric field?
Chemistry
1 answer:
stellarik [79]3 years ago
7 0

Answer:

3.1 x 10⁻²¹ Nm

Explanation:  

When placed in an external electric filed, an electric dipole experiences a torque. and this torque is represented mathematically with the equation:

torque (τ) = dipole moment vector (P) x electric field vector (E)

τ = P. E . sin θ

where θ is the angle between the water molecule and the electric field, which in this case is 90° (because this is where the torque is maximum)

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . sin90

τ = 6.2x10⁻³⁰Cm . 5.0x10⁸ N/C . 1

solve for τ

τ = 3.1 x 10⁻²¹ Nm

the maximum possible torque on the water molecule is therefore 3.1 x 10⁻²¹ Nm

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Solution:

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Number of moles = 1.4 mol

Now we will compare the moles of Reactant with product.

                 Zn         :          ZnCl₂

                  1           :             1

                 0.76     :           0.76

                Zn         :             H₂

                  1           :             1

                 0.76     :           0.76

               HCl         :          ZnCl₂

                  2           :             1

                 1.4         :           1/2×1.4 = 0.7

                HCl         :             H₂

                  2           :             1

                 1.4         :           1/2×1.4 = 0.7

Less number of moles of product are formed by HCl it will act limiting reactant.

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