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.
As we know that neutralization reaction is a reaction in which base react with acid to form salt and water.
When Potassium Hydroxide reacts with Sulphuric Acid, it forms Potassium Sulphate and Water.
As a result of neutralization reaction, Potassium Sulphate and Water is formed.
2KOH + H2SO4 ----> K2SO4 + 2H2O
Here, K2SO4 is found in aqueous medium in neutralization reaction. It is a neutral salt.
Answer:
S-type or silaceous asteroids are made up primarily of stony materials and nickel-iron. They inhabit the inner Asteroid Belt. M-type, or metallic, are made up mostly of nickel-iron, and are found in the middle region of the Asteroid Belt
Explanation:
I am going to assume that the reaction occurs in water since that is the easiest procedure (let me know if I am not supposed to make this assumption).
1) get 2 beakers, the chemicals for the reaction, DI water, and the catalyst.
2) equal volumes and amounts of the water was placed into the 2 beakers.
3) some of the catalyst was added to beaker the second beaker.
3) equal amounts of the chemicals for the reaction was placed into the 2 beakers.
4) the two reactions were compared to see how the un-catalyzed reaction in beaker 1 was effected by the introduction of a catalyst in beaker 2.
I hope this helps. Let me know in the comments if anything was unclear.
