1) mass composition
N: 30.45%
O: 69.55%
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100.00%
2) molar composition
Divide each element by its atomic mass
N: 30.45 / 14.00 = 2.175 mol
O: 69.55 / 16.00 = 4.346875
4) Find the smallest molar proportion
Divide both by the smaller number
N: 2.175 / 2.175 = 1
O: 4.346875 / 2.175 = 1.999 = 2
5) Empirical formula: NO2
6) mass of the empirical formula
14.00 + 2 * 16.00 = 46.00 g
7) Find the number of moles of the gas using the equation pV = nRT
=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)
=> n = 0.01769 moles
8) Find molar mass
molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol
9) Find how many times the mass of the empirical formula is contained in the molar mass
92.14 / 46.00 = 2.00
10) Multiply the subscripts of the empirical formula by the number found in the previous step
=> N2O4
Answer: N2O4
Answer: sorry I’m not sure
Odjri:
Answer:
4.78 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em></em>
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of MgSO₄ = 50.0 g,
mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.
mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>
Answer: heat
Other indicators: color change, odor, gas, temperature change.
