Virgin type pyrolysis Carbon black N330 System 1

Virgin type pyrolysis Carbon black N330 System 2

Virgin type pyrolysis Carbon black N330 System 3

Virgin type pyrolysis Carbon black N330

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Loading Port:: Qingdao

Payment Terms:: TT OR LC

Min Order Qty:: -

Supply Capability:: 20000 m.t./month

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Specifications

Carbon black
1.Black powder or granular pigment
2.N220.N330.N550,N660
3.Plastic,rubber industry
4.Hight quality

Carbon black

Introduction:

Our company maufactures the black canton series or superior-wear resistance(N200), resistance(N300),quick-press(N500),and common(N600) of dry process and wet process maily applied to various rubber products and PVC plastic.

Specifications:

Items	N220	N330	N550	N660
Indine absorption g/kg	121±5	82±5	43±4	36±4
DBP Absorption	114±5	102±5	121±5	90±5
CTAB Surface Area,	106-116	79-87	38-46	31-39
Nireogen Surface Area,	114-124	78-88	38-46	30-40
Tint Strength	110-120	98-108
Heating Loss,%,max	3.0	2.5	2.5	1.5
Ash,%,max	0.5	0.5	0.5	0.5
Tensile strength Msp,min,%	-0.8	-1.5	-3.5	-3.5
Exrension,%,min	+10	-10	-10	+10

Physical data:

Odor:None
PH:6-8
Boiling point :4200
Vapor pressuer:Not applicable
Solubility in water:Not applicable
Specific gravity:1.7-2.1
Viscosity: Not applicable

Packing:25kg knitting bags top grade linerfilm,good properties of waterproof and avoid water,solarization.

Virgin type pyrolysis Carbon black N330

Q: High school chemistry, catalyst activation energy map: Catalytic reaction is the reaction of the first reaction with the catalyst or attached to the catalyst to form intermediates, and then further reaction to produce products and catalysts, so the amount of catalyst in theory is the same! The activation of these two processes can be reduced! So there will be two peaks! Can be simplified as a peak!

Q: Why the catalyst after the chemical reaction of its quality and chemical properties unchanged: In fact, the catalyst reaction also enters, which changes the rate of positive and negative reactions ... enhances the reactive molecules of the reactants, thereby changing the reaction rate ...

Q: Cl + O3 ---&gt; ClO + O2O + ClO ---&gt; Cl + O2= O + O3 ----&gt; 2O2What is the catalyst? The intermediate?How do you know which is which? If the rate law is rate=k [O3] [Cl]determine:a) the overall order.b) unit for k.c) the rate determining step, justify your answer.: Cl is the catalyst. ClO the intermediate. The catalyst is the component which does not change in overall reaction. He forms some intermediate component(s) with the reactants. In the later reaction steps the intermediate(s) react forming the catalyst in its original state. (a) The overall order is the sum of the orders with respect to the components: n = 1 +1 = 2 (b) the unit of the rate of reaction is r [=] mol/ (Ls) (more general mol per unit time and volume) compare dimensions mol / (Ls) [=] k · mo/L · mol/L =k [=] L/(s mol) (more general unit volume per unit time and mole) (c) First reaction For elementary reaction steps the order of the reaction rate with respect to a reactant is equal to stoichiometric coefficient. Hence the rate of first reaction is: r? = k?·[Cl]·[O?] Overall rate is given by the rate determining step, while other reaction steps are in equilibrium: r = r? = k?·[Cl]·[O?] If second reaction is the rate determine step r? = k?·[O]·[ClO] while reaction 1 is at equilibrium K? = ( [ClO]·[O?] ) / ( [Cl]·[O?] ) =[ClO] = K?·( [Cl]·[O?] ) / [O?] the overall rate would be: r = r? = k?·[O]·[ClO] = K?·k?·[O]·[Cl]·[O?] / [O?] = k·[O]·[Cl]·[O?] / [O?] That doesn't match the observed rate law

Q: in my 99 ram 2500 v8 5.9 that code came up, not sure what to do about it or what it could be any ideas or help with be great thanks: This Site Might Help You. RE: catalyst efficiency low bank 1? in my 99 ram 2500 v8 5.9 that code came up, not sure what to do about it or what it could be any ideas or help with be great thanks

Q: okay im doing a project for my classroom about catalyst and i have to draw a picture but when i looked up on google i just saw a bunch of random stuff and a couple were metal so thats why im asking this question. :): a okorder /...

Q: High school stage which organic chemical reactions do not use catalyst: Aldehyde and silver ammonia solution reaction, and the new system of Cu (OH) 2 reaction.

Q: Does increasing the amount of catalyst added to, say, a solution of Hydrogen Peroxide, make the rate of reaction go faster. Is the rate of reaction directly proportional to the amount of catalyst added to the solution? Or does the experiment go at the same rate regardless of how much catalyst there is? Thanks would really appreciate some answers. - Sarah: Catalyst do not participate in the reaction but it speeds up the reaction. If you increase the amount of catalyst from the required amount, it would somehow negates its positive response or it would result in producing another product.

Q: What is the meaning of catalyst in chemistry?: In the chemical reaction can change the reaction rate of chemical reaction (increase or decrease) without changing the chemical balance, and its own quality and chemical properties in the chemical reaction before and after the material did not change the catalyst.

Q: How to poison the catalyst. What can be done?: In the reactants or catalyst mixed with a small amount of material, so that the catalyst catalytic capacity of a sharp decline or even loss, this phenomenon is called catalyst poisoning. For example, in the synthesis of ammonia feed gas containing CO, CO2 and H2S, PH3, water vapor and other impurities, can make iron catalyst poisoning; contact with the system of sulfuric acid, if arsenic and selenium oxide (As2O3, SeO2), can make vanadium catalyst Loss of activity. Therefore, it is necessary to purify the feed gas, prevent the poisoning of the catalyst, and also reduce the corrosion of the equipment. The phenomenon of catalyst poisoning is sometimes temporary, the removal of toxicants, the effectiveness of the catalyst can still be restored; sometimes it is permanent, without chemical treatment can not restore catalytic performance.

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