If the solubility of sugar at 25ºC is 211 g/100 g H2O, what mass of sugar can be dissolved in 300 g of H2O?
2 answers:
Answer:
633 grams of sugar can be dissolved in 300 g of H₂O
Explanation:
Solubility is the measure of the ability of a certain substance to dissolve in another and form a homogeneous system. Solubility is then the maximum amount of a solute that a solvent can receive and is expressed by concentration units.
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
Then:
You can apply the rule of three as follows: if by definition of solubility in 100 grams of H₂O there are 211 grams of sugar, in 300 g of H₂O how much sugar is there?
sugar= 633 grams
<u><em>633 grams of sugar can be dissolved in 300 g of H₂O</em></u>
Answer:
Explanation:
Hello,
In this case, solubility is defined as a condition wherein the amount of a solute has been completely dissolved in a certain amount of solvent and it is temperature-dependent as well. In such a way, for the given solubility at 25 °C, we understand that 211 g of sugar are completely dissolved in 100 g of water, therefore, for 300 g of water, we can compute the completely dissolved mass of sugar by using the following rule of three:
Best regards.
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Answer:
Only changes in temperature will influence the equilibrium constant . The system will shift in response to certain external shocks. At the new equilibrium
will still be equal to
, but the final concentrations will be different.
The question is asking for sources of the shocks that will influence the value of . For most reversible reactions:
- External changes in the relative concentration of the products and reactants.
For some reversible reactions that involve gases:
- Changes in pressure due to volume changes.
Catalysts do not influence the value of . See explanation.
Explanation:
.
Similar to the rate constant, the equilibrium constant depends only on:
the standard Gibbs energy change of the reaction, and
the absolute temperature (in degrees Kelvins.)
The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium the two processes balance each other. The concentration of each species will stay the same.
Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.
- Changes in concentration influence the number of particles per unit space.
- Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.
For reactions that involve gases,
- Changing the volume of the container will change the concentration of gases and change the reaction rate.
However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.