Color change, formation of a precipitation, formation of a gas, odor change, temperature change
We first consider the gases that will be present in that sample.
First, there will be nitrogen, as stated. Second, there will also be water in the form of water vapor. For this, we need the vapor pressure of water at 23.0 °C, which is about 21.0 mmHg. Now, the sum of the vapor pressures of the gases will be equivalent to the total pressure. So the pressure of nitrogen gas is:
785 - 21
= 764 mmHg
O3 + M2+(aq) + H2O(l) => O2(g) + MO2(s) + 2 H+
Eo(cell) = Eo(O3/O2) - Eo(MO2/M2+)
0.44 = 2.07 - Eo(MO2/M2+)
Eo(MO2/M2+) = 1.59 V
Answer:
NH₃
Explanation:
mass H = 6.10 grams
mass N = 28.00 grams
mass cpd = (6.10 + 28.00)grams = 34.10 grams
%H/100wt = (6.10/34.10)100% = 17.9% w/w
%N/100wt = (28.00/34.1)100% = 82.1% w/w
%/100wt => grams/100wt => moles => ratio => reduce => emp ratio
%H/100wt = 17.9% w/w => 17.9g => (17.9/1)moles = 17.9 moles H
%N/100wt = 82.1% w/w => 82.1g => (82.1/14)moles = 5.9 moles N
Ratio N:H => 17.9 : 5.9
Reduce mole ratio (divide by smaller mole value) => 17.9/5.9 : 5.9/5.9
=> 3HY:1H empirical ratio => empirical formula NH₃ (ammonia)
