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
Moles pf mgcl2= 2,11/M mgcl2=2,11/95= 0,0222
Molarity=0,0222/1,5=0,0148 M.
I hope this is correct.
Answer:
As Per Given Information
Number of moles SO₂ ( Sulphur dioxide ) is 0.30
We 've been asked to find the number of atoms in given moles of Sulphur dioxide .
Value of Avogadro Number is 6.022 × 10²³ mol⁻¹
For finding the number of atoms we will use formale
Putting the given value we obtain
So, the number of atoms in sulphur dioxide is 1.8066 × 10²³
I think it’s 44.6 J, but I’m not to sure so hoped this helped /:).
