From the equation, we can see that the molar ratio between hydrogen and oxygen is:
2 : 1
Next, we determine the moles of hydrogen and oxygen that are actually present using:
moles = mass / Mr
Hydrogen:
moles = 4 / 2 = 2
Oxygen:
10/32 = 0.3125
Therefore, it is evident that the moles of oxygen present, 0.3125, are less than those that are required for 2 moles of hydrogen, which is 1. This makes oxygen the limiting reactant, which is the one that limits the completion of a reaction.
C. Decreasing the temperature
D. Raising the pressure
<h3>Further explanation</h3>
Given
Reaction
2SO₂+O₂⇔2SO₃+energy
Required
Changes to the formation of products
Solution
The formation of SO₃ is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature decreases, the equilibrium will shift towards the exothermic reaction, so the reaction shifts to the right towards SO₃( products-favored)
And increasing the pressure, then the reaction shifts to the right SO₃( products-favored)⇒the number of coefficients is greater
Answer:
43.75 g of Nitrogen
Explanation:
We'll begin by calculating the mass of 1 mole of NH₄NO₃. This can be obtained as follow:
Mole of NH₄NO₃ = 1 mole
Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)
= 14 + 4 + 14 + 48
= 80 g/mol
Mass of NH₄NO₃ =?
Mass = mole × molar mass
Mass of NH₄NO₃ = 1 × 80 = 80 g
Next, we shall determine the mass of N in 1 mole of NH₄NO₃.
Mass of N in NH₄NO₃ = 2N
= 2 × 14
= 28 g
Thus,
80 g of NH₄NO₃ contains 28 g of N.
Finally, we shall determine the mass of N in 125 g of NH₄NO₃. This can be obtained as follow:
80 g of NH₄NO₃ contains 28 g of N.
Therefore, 125 g of NH₄NO₃ will contain = (125 × 28) / 80 = 43.75 g of N.
Thus, 125 g of NH₄NO₃ contains 43.75 g of Nitrogen
B) ampere
Ampere is apart of the metric unit so it would be reasonable for it to be a base unit.
I hope this helped!
