The mechanical-energy balance provides a basis for estimating pressure drop owing to friction in fluid flow. For steady flow of an incompressible fluid in a horizontal pipe of constant cross-sectional area, it may be written, ΔP ___ ΔL + 2 __ D f F rho u 2 = 0 where fF is the Fanning friction factor. Churchill13gives the following expression for fF for turbulent flow: f F = 0.3305 { ln [ 0.27 ∈ _ D + ( 7 _ Re ) 0.9 ] } −2 Here, Re is the Reynolds number and ∈ /D is the dimensionless pipe roughness. For pipe flow,Re ≡ urhoD/μ, where D is pipe diameter and μ is dynamic viscosity. The flow is turbulent for Re > 3000. Consider the flow of liquid water at 25°C. For one of the sets of conditions given below, determine m ˙ (in kg⋅s−1) and ΔP/ΔL (in kPa⋅m−1). Assume ∈ /D = 0.0001. For liquid water at 25°C, rho = 996 kg⋅m−3, and μ = 9.0 × 10−4 kg⋅m−1⋅s−1. Verify that the flow is turbulent. (a) D = 2 cm, u = 1 m·s−1
(b) D = 5 cm, u = 1 m·s−1
(c) D = 2 cm, u = 5 m·s−1
(d) D = 5 cm, u = 5 m·s−1

The mass has escaped into a gas which you can not see so the law stays true.

c. electrical charges of ionic bonds.

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Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.