Answer:
4.78 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em></em>
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of MgSO₄ = 50.0 g,
mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.
mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>
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This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:
Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:
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Answer:
Both B and D are correct.
Explanation:
B + H₂O ⇌ BH⁺ + OH⁻
If you add more products, the position of equilibrium will shift to the left to decrease their concentrations (Le Châtelier's Principle). The concentration of reactants will increase, but the equilibrium concentrations of products will also be higher than they were initially.
A is wrong. The equilibrium constant is a constant. It does not change when you change concentrations.
C is wrong. Per Le Châtelier's Principle, the concentrations must change when you ad a stress to a system at equilibrium.
(This is a poorly-worded question. "They" are probably expecting answer D.)
