Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
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hey there!:
density = 75.0 g/mL
Volume = 12 mL
mass = ?
Therefore:
D = m / V
75.0 =m / 12
m = 75.0 * 12
m = 900 g
Answer B
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We can find the mass of ammonia using the ideal gas law equation,
PV = nRT
where
P - pressure - 2.15 atm x 101 325 = 2.18 x 10⁵
V - volume - 3.00 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 15.0 °C + 273 = 288 K
substituting these values in the equation
2.18 x 10⁵ Pa x 3.00 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.273 mol
number of moles of NH₃ is 0.273 mol
molar mass of NH₃ - 17.0 g/mol
mass pf ammonia present - 0.273 mol x 17.0 g/mol = 4.64 g
mass of NH₃ present is 4.64 g
Answer:
No, The equation is not balanced
The correct ans is
2CO+O2=2CO2