35 b.
37 h
39 i
41 d
hope that helps
1) Balanced chemical equation:
2SO2 (g) + O2 (g) -> 2SO3 (l)
2) Molar ratios
2 mol SO2 : 1 mol O2 : 2 mol SO3
3) Convert 6.00 g O2 to moles
number of moles = mass in grams / molar mass
number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.
4) Use proportions with the molar ratios
=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2
=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.
5) Convert 0.375 mol SO2 to grams
mass in grams = number of moles * molar mass
molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol
=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g
Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.
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