Mass = 5.90 x 114 = 672.6g.
This can be rounded to <em>673g</em>
<em>C = 12 grams x 8 = 96 grams </em>
<em>H = 1 gram x 18 = 18 grams </em>
<em>Total = 96 + 18 = 114 grams in one mole. </em>
<u>If you have 5.9 moles, multiply this by the weight of one mole (114 grams) = 672.6 rounded to 673 grams in 5.9 moles. </u>
True!! they can be measured without changing chemical identity.
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Answer:
5.36 grams the mass in grams of zinc nitrate the chemist has added to the flask.
Explanation:
Moles of zinc nitrate = n
Volume of the solution = 135.0 mL = 0.1350 L
Molarity of the solution = 0.21 M
Mass of 0.02835 moles of zinc nitrate:
0.02835 mol × 189 g/mol = 5.358 g ≈ 5.36 g
5.36 grams the mass in grams of zinc nitrate the chemist has added to the flask.
1) Convert 12.9 liters of Oxygen to mol at the given conditions:
PV = nRT ⇒ n = PV/RT
n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]
n = 0.636 mol of O2
2) use the stoichiometry derived from the balanced chemical equation
1mol C2H4 / 3 mol O2 = x mol C2H4 / 0.636 mol O2
x = 0.636 / 3 mol O2 = 0.212 mol O2.
Answer: 0.212 mol O2
