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filtration    音标拼音: [fɪltr'eʃən]
n. 过滤,筛选

过滤,筛选

filtration
n 1: the process whereby fluids pass through a filter or a
filtering medium
2: the act of changing a fluid by passing it through a filter

Filtration \Fil*tra"tion\, n. [Cf. F. filtration.]
The act or process of filtering; the mechanical separation of
a liquid from the undissolved particles floating in it.
Fimble


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  • Questions asked by - Socratic
    What is the difference between gravity filtration and vacuum filtration? When should vacuum filtration be used? What is the difference between copper (II) sulphate trihydrate, copper (II) sulphate pentahydrate, and anhydrous copper (II) sulphate? Methane, CH4 , the major component of natural gas burns in air to form CO2 and H2O
  • Question #7e2d2 - Socratic
    During filtration, some of the particles pass through the filter while others, being larger, are trapped But no molecules are changed in structure in any way, which is the primary requirement of a chemical change
  • Site Map - Separating Mixtures Questions and Videos | Socratic
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  • Solve the following quadratic equation using factoring? x²+5x+6=0
    Please see below 1 x²+5x+6=0 can be written as x^2+3x+2x+6=0 or x(x+3)+2(x+3)=0 or (x+2)(x+3)=0 i e either x+2=0 means x=-2 or x+3=0 means x=-3 x²=4x-5 can be
  • Who were the Axis powers of WWII? What areas did they . . . - Socratic
    Japan-Italy-Germany They controlled most of Europe and of the Pacific France had been invaded, Denmark and Norway had been submitted Only the UK was resisting while Switzerland and Sweden had remained neutral The Pacific was the realm of the Japanese, they controlled the Eastern coast of China, Korea, only Hawaii and Ocenania were not under their yoke
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    y = -x^2 - 2x + 8 1 Answer Nghi N May 21, 2015 #y = -x^2 - 2x + 8# Answer link You can reuse this answer
  • Question #cbd67 - Socratic
    Explanation: Note that the denominator is: #x^2+2x+1 = (x+1)^2# so we can substitute #t=x+1#, #x=t-1#, #dx=dt# and have: #int (x^3dx) (x^2+2x+1) = int ( (t-1)^3dt) t
  • How do you proof lim x → ∞ (n^ (x+1)) ( (x+1)!) = 0 - Socratic
    below Assuming the Stirling approx is allowed, ie: ln Q! approx Q ln (Q) - Q lim_(x to oo) (n^(x+1)) ((x+1)!) let z = x+1 = lim_(z to oo) exp( ln( (n^z) (z!))) = lim





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