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What is the percent yield of O_2 if 10. 2 g of O_2 is . . . - Socratic %Yield=67 5% The percent yield can be calculated by: %Yield=("Actual Yield") ("Theoretical Yield")xx100% The actual yield of oxygen is given and it is 10 2g We will need to find the theoretical yield The decomposition reaction of water can be written as: 2H_2O(l)->2H_2(g)+O_2(g) To find the theoretical yield of oxygen we will use dimensional analysis: ?g O_2=underbrace(17 0gH_2Oxx(1molH_2O
Please can someone help and balance this equation [Ni(H2O)6 . . . - Socratic Well, the first is a chemical equation, the which I am competent to address We make the hexa-ammine nickel(II) complex [Ni(OH_2)_6]^(2+) + 6NH_3(aq) rarr [Ni(NH_3)_6]^(2+) + 6H_2O(l) And the second is a nucular reaction the which DOES NOT conserve mass ""^235U + "neutron" rarr ""^140Xe + ""^94Sr + "2 neutrons " And please compare this with your text, and notes because I am no
CH4 + 2O2 - gt; CO2 + 2H2O + heat If 5 mole of methane (CH4 . . . - Socratic O_2 = 10 moles FIRST we make sure that the equation is balanced Then we compare the coefficients of the compounds to find the molar ratios of reaction CH_4 + 2O_2 -> CO_2 + 2H_2O is already balanced, so we only need to look at the ratio of methane and oxygen CH_4 : O_2 = 1:2 or O_2 = 2xxCH_4 So, for 5 moles of methane we obtain: O_2 = 2xx5 = 10 moles
How do the carbon, nitrogen, hydrologic, and phosphorus . . . - Socratic Simple answer: through life processes such as respiration, elimination of body wastes, and death of organisms that return C, N, H2O and P to the environment All organisms are naturally a part of these 4 cycles, and as such return or revert some compounds to the environment due to daily life processes such as respiration, photosynthesis, excretion of wastes, etc In addition, once an organism
What is the oxidizing agent in the following net ionic . . . - Socratic See below Oxidizing agent is the species that is reduced, being reduced means to gain electrons and have a lowered oxidation number Let's see: I^- in the reactants to I^(+0), increased oxidation state therefore it was oxidized Na^+ to Na^+ no change S^(+6) in the reactants to S^(+4), decreased oxidation state therefore it was reduced, so the sulfate ion acted as the oxidizing species by being
Why does [Co (NN_3)_6]^ (3+) form an inner orbital complex but . . . sf([Co(NH_3)_6]^(3+)) is an inner orbital complex because it adopts a low spin electronic configuration I am assuming you mean sf(NH_3) as the ligand in question in the 1st example One characteristic of the transition elements is their ability to form complex ions These consist of a central metal ion surrounded by electron - donating species called ligands Cobalt(III) ions have the