Answer:
Answer: The solubility of B is high than the solubility of A.
Explanation:
The solubility is defined as the amount of substance dissolved in a given amount of solvent. More the solute gets dissolved, high will be the solubility and less the solute dissolved, low will be the solubility.
Mass of undissolved substance of substance A is more than Substance B at every temperature. This implies that less amount of solute gets dissolved in the given amount of solvent.
Therefore, B has high solubility than substance A.
Answer:
The mass of water to be added is 2 pounds
Explanation:
The given parameters are;
The mass of the given solution = 2 pounds
The concentration of the given solution = 30%
The desired concentration of the solution = 15%
The mass, m of the acetic acid in the given solution = 30% × 2 pounds
m = 30/100 × 2 pounds = 0.6 pounds
To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;
15% of X = 0.6 pounds
15/100 × X = 3/20 × 0.6 pounds
∴ The mass of the solution required X = 0.6 × 20/3 = 4 pounds
The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds
Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.
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
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