Answer:
0.0375 moles HBr
Explanation:
we are given;
- Molarity of HBr solution as 0.15 M
- Volume of the solution as 250 mL
We are required to determine the number of moles;
We need to know that;
Molarity = Moles ÷ Volume
Therefore;
Moles of HBr = Molarity of HBr solution × Volume of solution
Thus;
Moles of HBr = 0.15 M × 0.25 L
= 0.0375 Moles
Thus, the number of moles of HBr solute is 0.0375 moles
A high concentration of water has <u>fewer</u> dissolved particles than a low water concentration.
Most cell membranes are not as easily permeable to many dissolved compounds as water is. There is a quick and constant flow of water. From one area with less dissolved matter to another with more, water transports NET. Or, if you want, from an area with a lot of water to one with little water. The terms isotonic, hypotonic, and hypertonic refer to the concentration of dissolved material. In a medium, such as the extracellular fluid, every distinct material has a concentration gradient that is unique from the gradients of other substances. Every substance will diffuse in line with that gradient as well.
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Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
