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Anonymous users2024-02-08The concentration is on the low side. When the liquid in the beaker is transferred to the volume, a portion of the solution remains in the beaker and on the glass rod, which must be washed with distilled water several times before being transferred to the volumetric flask to minimize losses.
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Anonymous users2024-02-07Low step 3 beakers and glass rods should be washed 2 3 times with clean water and the washing solution should be added to the volumetric flask.
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Anonymous users2024-02-06According to the title, 1 mol of mixed gas is fully burned to produce 2 mol of CO2 and molH2O, so 1 mol of mixed gas contains 2 mol of carbon atoms and mol. of hydrogen atoms
The molecular formula of ethylene is C2H4, so there are also 2 carbon atoms in the molecule of another hydrocarbon A, and the hydrogen atom is greater than 4, and only C2H6 is eligible
Therefore, the molecular formula of a is C2H6
Suppose that 1 mol of mixed gas contains x mol of ethylene, and the amount of ethane is (1-x)mol, then 4x + 6(1-x)=,x = , that is, 1 mol of mixed gas contains mol of ethylene and mol ethane.
The mass fraction of ethylene is, and the mass fraction of ethane is.
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Anonymous users2024-02-05C2H4 + 2O2 = Ignition = 2CO2 + 2H2O
Assuming that 2mol of gas is all ethylene, 4molCO2 and 4molH2O are generated, because 2mol of ethylene and 2mol of another hydrocarbon A gas mixture are burned, and 4molCO2 is formed
Therefore, the number of carbon atoms contained in another hydrocarbon must also be 2, that is, the chemical formula is C2Hx, and it can only be C2H2, or C2H6
Because the volume of water produced after combustion is larger than the volume of water produced by all gases produced by ethylene, it can be determined that the other gas must be C2H6, 2C2H6 + 7O2 = ignition = 4CO2 + 6H2O
Assuming that all the gases are C2H6, then 2molC2H6 produces 6molH2O
Because it is generated at this time, it is crossed:
Quantity of C2H4 substance: Amount of C2H6 substance = (:(3, after which its mass fraction can be calculated.
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Anonymous users2024-02-04Let cxhy2cxhy + o2 ==== 4CO2 + H2O get: x=2 y=
Therefore, the other hydrocarbon is C2H6, that is, ethane (only the amount ratio of H supersubstances in ethane in C2: N(C2H4):N(C2H6)=3:
2 Mass ratio: m(c2h4):m(c2h6)=(3 28):
2naoh+so2=na2so3+h2o naoh+so2=nahso3
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Generally speaking, an outer electron number of 8 is a stable structure, and atoms have a tendency to make their outermost electrons become 8-electron stable structure. The X element with an outer electron number of 3 can gain 5 electrons or lose 3 electrons (the subouter shell is generally 8 electrons) can make itself a stable structure with the outermost shell of 8 electrons, because it is much more difficult to get 5 electrons than to lose 3 electrons, so it often loses the outermost 3 electrons and becomes a stable structure with the outermost 8 electrons, showing a valency of +3 valence. >>>More
Alkaline Therefore, c(oh-)>c(h+) a is wrong.
Soluble in water alkaline, it is its weak acid S2 - step by step hydrolysis and the production of OH- makes the solution alkaline, these OH- all come from water, and only for the aqueous solution, there is C (OH-) = C (H+) and H+ in the solution in the form of Hs- H2S H+, so option B: >>>More