Answer: the original volume was 0.98L
Explanation:
V1 =?
T1 = 22k
V2 = 3.8L
T2 = 85k
V1 /T1 = V2 /T2
VI/22 = 3.8/85
V1 = 22 x (3.8/85)
V1 = 0.98L
Answer: 3 <span>moles of water would be produced in present case.
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Reason:
Reaction involved in present case is:
<span> C5H12 + 8O2 </span>→<span> 5CO2 + 6H2O
In above reaction, 1 mole of C5H12 reacts with 8 moles of oxygen to give 6 moles of water.
Thus, 4 moles of oxygen will react with 0.5 mole of C5H12, to generate 3 moles of H2O.</span>
Answer:
The answer to your question is: C₂H₆O₁ = C₂H₆O
Explanation:
Data
CxHyOz
mass = 40 g produced 76.40 g of CO2
46.96 g of H2O
Empirical formula = ?
Process
CxHyOz + O2 ⇒ CO2 + H2O
44g of CO2 -------------------- 12 g of Carbon
76.40 g of CO2 --------------- x
x = 20.84 g of Carbon
12 g of Carbon --------------- 1 mol
20.84 g of C --------------- x
x = (20.84 x 1) / 12
x = 1.74 mol of Carbon
18 g of H2O -------------------- 2 g of H
46.96 g of H2O -------------- x
x = (46.96 x 2) / 18
x = 5.22 g of H
1 g of H ------------------------ 1 mol of H
5.22 g of H ------------------- x
x = 5.22 mol of H
Mass of Oxygen = 40 - 20.84 - 5.22g
= 13.94 g
16 g of Oxygen ---------------- 1 mol
13.94 g of O -------------------- x
x = 0.87 mol of O
Divide by the lowest number of moles
Carbon = 1.74 / 0.87 = 2
Hydrogen = 5.22 / 0.87 = 6
Oxygen = 0.87 / 0.87 = 1
Empirical formula
C₂H₆O₁ = C₂H₆O
The sample is most likely lead with lighter substances in it. Pure lead would be closer to the pure lead density, and heavier substances would make the density more, not less.
Answer:
Explanation:
0.5678 G X GRAMS
KHC8H4O4 + NaOH = NaKC8H4O4 + H2O
1 MOL 1 MOL
0.5678G X 204G/MOL = 0.00278 MOL KHC8H4O4
0.00278 MOL KHC8H4O4 X 1 MOLE NaOH/1 MOLE KHC8H4O4=0.00278 MOL NaOH
0.00278 MOL NaOH/26.26ml=0.106 molar