The reaction involved in present case is:
Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)
In terms of oxidation and reduction, the reaction can be shown at
Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)
From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.
From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.
Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.
1 mole of any gas occupy 22.4 L at STP (standard temperature and pressure, 0°C and 1 atm).
Let given gases be 1 mole. So their volumes will be the same, 22.4 liters.
Density is the ratio of mass to volume.
By formula; density= mass/volume; d=m/V
To find out masses of gases, do the mole calculation.
By formula; mole= mass/molar mass; n= m/M; m= n*M
Molar masses are calculated as
1. C₂H₆ (ethane) = 2*12 g/mol + 6*1 g/mol= 30 g/mol
2. NO (nitrogen monoxide) = 1*14 g/mol + 1*16 g/mol= 30 g/mol
3. NH₃ (ammonia) = 1*14 g/mol + 3*1 g/mol= 17 g/mol
4. H₂O (water) = 2*1 g/mol + 1*16 g/mol= 18 g/mol
5. SO₂ (sulfur dioxide) = 1*32 g/mol + 2*16 g/mol= 64 g/mol
Use Periodic Table to get atomic mass of elements.
Since their volumes are equal, compounds having the same molar mass will have the same density.
Recall the formula d= m/V.
Ethane and nitrogen monoxide have the same density.
The answer is C₂H₆ and NO.
Answer:
616,0 ng is the right answer.
Explanation:
You should know that 1 mole = 1 .10^9 nanomoles
Get the rule of three.
1 .10^9 nanomoles ...................... 56.0 gr
11 nanomoles .....................
(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr
Let's convert
6.16 x 10^-7 gr x 1 .10^9 = 616 ngr
Answer:
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