Is bubble chamber one of your choices? Bubble chamber sounds like a good fit for the question.
At STP one mole of any gas occupies 22.4 L
moles NO2 = 99.0/22.4 = 4.42
mass NO2 = 4.42 mol x 46.0 g/mol=203 g
Answer: Rate in terms of disappearance of
= ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
Rate in terms of disappearance of
= ![-\frac{1d[Cl_2]}{1dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BCl_2%5D%7D%7B1dt%7D)
Rate in terms of appearance of
= ![\frac{1d[NOCl]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNOCl%5D%7D%7B2dt%7D)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of = ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
Rate in terms of disappearance of = ![-\frac{1d[Cl_2]}{1dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BCl_2%5D%7D%7B1dt%7D)
Rate in terms of appearance of
= ![+\frac{1d[NOCl]}{2dt}](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BNOCl%5D%7D%7B2dt%7D)
the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ
There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.
Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).
Thus, 24.3 g of benzene contains:
n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.
2 moles = 6278 kJ
0.311 moles =x
By the straightforward direct three rule:
2x = -1953.08 x = -976.5 kJ
Learn more about combustion here-
brainly.com/question/15117038
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