Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present.
Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ
Explanation:
Using this formular, q = [mCpΔT] and = [nΔHfusion]
The energy that is needed in the different physical changes is thus:
The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:
q = [mCpΔT]
q = 52.0 x 2.09 x 10
q = 1.09 kJ
While from 0°C to 100°C is calculated as:
q = [mCpΔT]
q = 52.0 x 4.18 x 100
q = 21.74 kJ
And for fusion at 0°C is called Heat of fusion and would be given as:
q = n ΔHfusion
q = 52.0 / 18.02 x 6.02
q = 17.38 kJ
And that required for vaporization at 100°C is called Heat of vaporization and it's given as:
q = n ΔHvaporization
q = 52.0 / 18.02 x 40.7
q = 117.45 kJ
Add up all the energy gives 157.8 kJ
Just to make sure I’m right, is number 1 miss spelled??
