439.3 g CO2
Explanation:
First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:
3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)
= 9.981 mol CO2
Use the molar mass of CO2 to determine the # of grams of CO2:
9.981 mol CO2 x (44.01 g CO2/1 mol CO2)
= 439.3 g CO2
Am - it has an atomic number of 95 which is greater than 92.
Transuranium elements are elements with atomic levels greater than 92
Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol
1 mole ---------- 84 g
? mole ---------- 110 g
moles NaHCO₃ = 110 . 1 / 84
moles NaHCO₃ = 110 / 84
= 1.309 moles
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Answer A
x-ray optical telescope.
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