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Work=Force*Displacment
Work=.05N * .07m = .0035J
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37.1×10−8M is the concentration of pb 2 ions in a solution prepared by adding 5. 00 g of lead(ii) iodide to 500. ml of 0. 150 m ki? [ k sp(pbi 2) =1. 4 × 10 –8] by common ion effect.
A phenomenon known as the "common ion effect" allows for the modification of a salt's molar solubility by adding another salt in which one ion dissociates in solution and maintains equilibrium with the undissociated salt.
The term "common-ion effect" describes the reduction in solubility of an ionic precipitate caused by the addition of a soluble molecule that shares an ion with the precipitate to the solution. Le principle for the equilibrium response of ionic association/dissociation leads to this behavior.
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C The number of outermost electrons increases.
Answer:
D. ![K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5B%5Ctext%7BNO%7D_%7B2%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BHNO%7D_%7B2%7D%5D%7D)
Explanation:
The general form of an equilibrium constant expression is
![K = \frac{[\text{Products}]}{[\text{Reactants}]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5B%5Ctext%7BProducts%7D%5D%7D%7B%5B%5Ctext%7BReactants%7D%5D%7D)
In the equilibrium
HNO₂ ⇌ H⁺ + NO₂⁻
The products are H⁺ and NO₂⁻, and the reactant is HNO₂.
∴
The silver ion concentration in saturated solution of silver (i) phosphate is calculated as follows.
write the equation for dissociation of silver (i) phosphate
that is Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)
let the concentration of the ion be represented by x
ksp is therefore= (3x^3 )(x) = 1.8 x10 ^-18
27 x^3 (x) = 1.8 x10^-18
27x^4 = 1.8 x10^-18 divide both side 27
X^4 = 6.67 x10 ^-20
find the fourth root x = 1.6 x10 ^-5m
the concentration of silver ion is therefore = 3 x (1.6 x10^-5) = 4.8 x10^-5m
