Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
Answer:
H(aq) + NO3 (aq) + HF(aq)
Explanation:
In the given mixture of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water the major species present are H(aq) + NO3 (aq) + HF(aq).
On the reaction of HNO3 (Nitric Acid) and HF (hydrofluoric acid) in water , it will give a polar solution and will form a homogenous mixture.
Hence, the correct answer is "H(aq) + NO3 (aq) + HF(aq)".
Answer:
<u>Balanced equation:</u>

Explanation:
The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.


<u>Ionic equation:</u>

Cancel the same ions on the both sides of the reaction.
The net ionic equation is as follows.

Answer:
B
Explanation:
by simlfying it you get HCO2