Answer:
0.069moles
45.65%
Explanation:
Firstly, we need to calculate the number of moles of CO2 produced. We can use the ideal gas equation for this.
PV = nRT
n = PV/RT
according to the question,
P = 746torr
V = 1.73L
T = 26 = 26 + 273.15 = 299.15K
n = ?
R = 62.364 L. Torr/k.mol
n = (746 * 1.73)/(62.364 * 299.15) = 0.069moles
B. To get this, we can use their molar masses. The molar mass of calcium carbonate is 100g/mol while for magnesium carbonate, molar mass is 84g/mol
The percentage by mass is (84)/(84 + 100) * 6.53g = 2.98g
= 2.98/6.53 * 100 = 45.65%
Answer:
2.66 g of Fe, can be obtained from the reaction
Explanation:
Let's think the reaction:
2Fe₂O₃ + 6CO → 4Fe + 6CO₂
Ratio is 2:4, so If i have x moles of iron (III) oxide, I will produce the double of moles of Fe.
Mass / Molar mass = Mol
3.80 g / 159.7 g/m = 0.0237 moles
0.0237 moles . 2 = 0.0475 moles
Molar mass Fe = 55.85 g/m
Mol . Molar mass = Mass → 0.0475 m . 55.85 g/m = 2.66 grams
Explanation:
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