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
Answer:
Explanation:
Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.
In 1897, J.J. Thomson discovered the electron by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.
Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny and heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.
The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons. Bohr was the first to discover that electrons travel in separate orbits around the nucleus and that the number of electrons in the outer orbit determines the properties of an element.
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