倍比稀释英文怎么说
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[倍比稀释英文怎么说] 请教各位，倍比稀释英文应该怎么说啊？谢谢！ 关键词:[]…
请教各位，倍比稀释英文应该怎么说啊？谢谢！
回复倍比(倍数比率)： multiple proportion稀释： dilution只查到几个单词，英文不好，希望没有误导您。回复gradual dilution回复如果是双倍的系列稀释，可以用doubling dilution， 如果是更高倍数的系列稀释，可以用serial dilution, 希望有帮助。回复2-fold dilution回复gradual dilution 是比较地道的说法。回复比较地道的翻译:倍比稀释:
2-fold dilution 连续倍比稀释:
serial 2-fold dilution 连续10倍稀释:
serial 10-fold dilution回复谢谢各位的帮助了，十分感谢！回复学到一招，happy!谢谢各位！
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频道总排行
频道本月排行effluent dilution 的翻译是:流出稀释 isotopic dilution 的翻译是:同位素稀释 of or relating to a solution whose dilution has been reduced. 的翻译是:或与解答相关稀释减少了。 stable isotope dilution mass spectrometry 的翻译是:稳定同位素稀释质谱分析 heat of dilution 的翻译是:冲淡热 stable-isotope dilution technique 的翻译是:稳定同位素稀释技术 the right granting to shareholders the first opportunity to buy
provides protection against dilution of the shareholder's ownership interest. 的翻译是:正确授予股东第一个机会买股票的一个新问题;提供防护反对股东的所有者权益的稀释。 down-hole dilution 的翻译是:向下钻进稀释 dilution of dust air 的翻译是:尘土空气的稀释 dilution air plant 的翻译是:稀释压缩空气装置 serial dilution 的翻译是:连续稀释 Ostwald dilution law 的翻译是:Ostwald稀释法律 hydrogen and ascorbate dilution determination 的翻译是:氢和抗坏血酸稀释决心 sample probe for dilution air 的翻译是:稀释空气的试样探针 sample collection bag for dilution air 的翻译是:样品稀释空气的汇集袋子 limit dilution passage 的翻译是:极限稀释段落 dilution analysis 的翻译是:稀释分析 dilution bottle 的翻译是:稀释瓶 dilution cloning 的翻译是:稀释克隆 dilution coefficient 的翻译是:稀释系数
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Copyright (C) 2005 - . All Rights Reserved.serial dilution是什么意思_百度知道
serial dilution是什么意思
serial dilution英[ˈsiəriəl daɪˈlu:ʃən]美[ˈsɪriəl daɪˈluʃən][释义] 连续稀释;例句：The sensitivity of assay was tested with ten fold serial dilution of CDV and compared with ordinary RT-PCR.利用十倍稀释法检验方法的灵敏度并与普通RT-PCR法进行比较；Methods: MIC was determined by microdilution method.BBF was established by plate culture method and defined by being stained with AgNO3.Viable bacteria were counted by serial dilution.方法：采用平板法培养细菌生物被膜，微量稀释法测定抗菌药物的最低抑菌浓度(MIC)，银染法快速鉴定BBF，连续稀释法进行活菌计数。
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出门在外也不愁Resource Materials: Making Simple Solutions and Dilutions
| | concentrated stock solutions (X units) | normality - molarity conversion
Simple Dilution (Dilution Factor Method based on ratios)
A simple dilution is one in which
a unit volume of a liquid material of interest is combined
with an appropriate volume of a solvent liquid to achieve
the desired concentration. The dilution factor is the
total number of unit volumes in which your material will be dissolved.
The diluted material must then be thoroughly mixed to achieve
the true dilution. For example, a 1:5 dilution (verbalize as
&1 to 5& dilution) entails combining 1 unit volume
of solute (the material to be diluted) + 4 unit
volumes of the solvent medium (hence, 1 + 4 = 5
= dilution factor). The dilution factor is frequently expressed
using exponents: 1:5 would be 5e-1; 1:100 would be 10e-2, and
Example 1: Frozen orange juice concentrate is usually diluted
with 4 additional cans of cold water (the dilution solvent) giving
a dilution factor of 5, i.e., the orange concentrate represents
one unit volume to which you have added 4 more cans (same unit
volumes) of water. So the orange concentrate is now distributed
through 5 unit volumes. This would be called a 1:5 dilution,
and the OJ is now 1/5 as concentrated as it was originally. So,
in a simple dilution, add one less unit volume of solvent
than the desired dilution factor value.
Example 2: Suppose you must prepare 400 ml
of a disinfectant that requires 1:8 dilution from a concentrated
stock solution with water. Divide the volume needed by the dilution
factor (400 ml / 8 = 50 ml) to determine the unit volume. The
dilution is then done as 50 ml concentrated disinfectant + 350
Serial Dilution
A serial dilution is simply a series
of simple dilutions which amplifies the dilution factor quickly
beginning with a small initial quantity of material (i.e., bacterial
culture, a chemical, orange juice, etc.). The source of dilution
material (solute) for each step comes from the diluted material
of the previous dilution step. In a serial dilution the total
dilution factor at any point is the product of the
individual dilution factors in each step leading up to it.
Final dilution factor
(DF) = DF1 * DF2 * DF3
Example: In a typical microbiology exercise the students
perform a three step 1:100 serial dilution of a bacterial
culture (see figure below) in the process of quantifying the
number of viable bacteria in a culture (see figure below). Each
step in this example uses a 1 ml total volume. The initial step
combines 1 unit volume of bacterial culture (10 ul) with 99 unit
volumes of broth (990 ul) = 1:100 dilution. In the second step,
one unit volume of the 1:100 dilution is combined with
99 unit volumes of broth now yielding a total dilution of 1:100x100
= 1:10,000 dilution. Repeated again (the third step) the total
dilution would be 1:100x10,000 = 1:1,000,000 total dilution.
The concentration of bacteria is now one million times less
than in the original sample.
Making fixed volumes of specific concentrations from liquid reagents:
V1C1=V2C2 Method
Very often you will need to make a specific
volume of known concentration from stock solutions, or perhaps
due to limited availability of liquid materials (some chemicals
are very expensive and are only sold and used in small quantities,
e.g., micrograms), or to limit the amount of chemical waste.
The formula below is a quick approach to calculating such dilutions
V = volume, C = in whatever units
you are working.
(stock solution attributes) V1C1=V2C2 (new
solution attributes)
Suppose you have 3 ml of a stock solution of 100 mg/ml ampicillin
(= C1) and you want to make 200 ul (= V2) of
solution having 25 mg/ ml (= C2). You need to
know what volume (V1)
of the stock to use as part of the 200 ul total volume needed.
V1 = the volume of stock you will start with. This
is your unknown.
C1 = 100 mg/ ml in the stock solution
V2 = total volume needed at the new concentration
= 200 ul = 0.2 ml
C2 = the new concentration = 25 mg/ ml
By algebraic rearrangement:
V1 = (V2 x C2)
ml x 25 mg/ml) / 100 mg/ml
and after cancelling the
V1 = 0.05 ml, or 50 ul
So, you would take 0.05 ml = 50 ul
of stock solution and dilute it with 150 ul of solvent to
get the 200 ul of
25 mg/ ml solution needed. Remember that the amount of solvent
used is based upon the final volume needed, so you have to subtract
the starting volume form the final to calculate it.
Moles and Molar solutions (unit = M = moles/L)
Sometimes it may be more efficient to
use molarity when calculating concentrations. A mole
is defined as one gram molecular weight of an element or compound,
and comprised of exactly 6.023 x 10^23 atoms or molecules (this
is called Avagadro's number). The mole is therefore a unit expressing
the amount of a chemical. The mass (g) of one mole
of an element is called its molecular weight (MW).
When working with compounds, the mass of one mole of the
compound is called the formula weight (FW). The distinction
between MW and FW is not always simple, however, and the terms
are routinely used interchangeably in practice. Formula (or molecular)
weight is always given as part of the information on the label
of a chemical bottle.
The number of moles in an arbitrary mass
of a dry reagent can be calculated as:
# of moles = weight (g)/ molecular weight (g)
is the unit used to describe the number of moles of a chemical
or compounds in one liter (L) of solution and is thus a unit
of concentration. By this definition, a 1.0 Molar (1.0
M) solution is equivalent to one formula weight (FW =
g/mole) of a compound dissolved in 1 liter (1.0 L) of solvent
(usually water).
Example 1: To prepare a liter of a simple
molar solution from a dry reagent
Multiply the formula
weight (or MW) by the desired molarity to determine how many
grams of reagent to use:
Chemical FW =
194.3 g/ to make 0.15 M solution use
194.3 g/mole * 0.15 moles/L = 29.145 g/L
Example 2: To prepare a specific volume of a specific
molar solution from a dry reagent
A chemical has a FW of 180 g/mole and
you need 25 ml (0.025 L) of 0.15 M (M = moles/L) solution. How
many grams of the chemical must be dissolved in 25 ml water to
make this solution?
#grams/desired volume (L) = desired molarity
(mole/L) * FW (g/mole)
by algrebraic rearrangement,
#grams = desired volume (L) * desired molarity
(mole/L) * FW (g/mole)
#grams = 0.025 L * 0.15 mole/L * 180 g/mole
after cancelling the units,
#grams = 0.675 g
So, you need 0.675 g/25 ml
For more on molarity, plus molality and
normality:
More examples of worked problems:
Percent Solutions (% = parts per hundred or grams/100 ml)
Many reagents are mixed as percent
concentrations as weight per volume for dry reagent OR volume
per volume for solutions. When working with a dry reagent
it is mixed as dry mass (g) per volume and can be simply
calculated as the % concentration (expressed as a proportion
or ratio) x volume needed = mass of reagent to use.
Example 1: If you want to make 200 ml of 3 % NaCl you would
dissolve 0.03 g/ml x 200 ml = 6.0 g NaCl in 200 ml water.
When using liquid reagents the
percent concentration is based upon volume per volume,
and is similarly calculated as % concentration x volume needed
= volume of reagent to use.
Example 2: If you want to make 2 L of 70% actone you would
mix 0.70 ml/ml x 2000 ml = 1400 ml acetone with 600 ml water.
To convert from % solution to molarity, multiply the % solution by 10 to express the
percent solution grams/L, then divide by the formula weight.
Molarity =
(grams reagent/100 ml) * 10
xxxxxxxxxxFW
Example 1: Convert a 6.5 % solution of a chemical with
FW = 325.6 to molarity,
[(6.5 g/100 ml) * 10] / 325.6
g/mole = [65 g/L] / 325.6g/mole = 0.1996 M
To convert from molarity
to percent solution,
multiply the molarity by the FW and divide by 10:
% solution
= molarity * FW
xxxxxxxxxx10
Example 2: Convert a 0.0045 M solution of
a chemical having FW 178.7 to percent solution:
[0.0045 moles/L
* 178.7 g/mole] / 10 = 0.08
% solution
6. Concentrated
stock solutions - using &X& units
Stock solutions of stable
compounds are routinely maintained in labs as more concentrated
solutions that can be diluted to working strength when used in
typical applications. The usual working concentration is denoted
as 1x. A solution 20 times more concentrated would be denoted
as 20x and would require a 1:20 dilution to restore the typical
working concentration.
Example: A 1x solution of a compound has a molar concentration
of 0.05 M for its typical use in a lab procedure. A 20x stock
would be prepared at a concentration of 20*0.05 M = 1.0 M. A
30X stock would be 30*0.05 M = 1.5 M.
7. Normality
(N): Conversion to Molarity
Normality = n*M where
n = number of protons (H+) in a molecule of the acid.
Example: In the formula for concentrated sulfuric (36
N H2SO4), there are two protons, so, its molarity= N/2. So, 36N
H2SO4 = 36/2 = 18 M.
Modified 9-27-12 ga
Lewiston, ME 04240}