If 11.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
1 answer:
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Answer:
The steps are explained below, the essential step is to find mass here, 120 g of NaOH.
Explanation:
In order to answer this question, we need to define molarity conceptually firstly to see what variables we need. According to the formula, molarity is equal to the ratio between moles and volume, while moles itself is a ratio between mass and molar mass. This means we have a formula for molarity involving mass, molar mass and volume:
In order to prepare a 500.0 mL of stock solution of 6.0 M of NaOH, we then need to find the mass of NaOH dissolved in this solution using the equation above:
Now, since we have the mass of NaOH, we can describe the steps needed to prepare this solution:
- measure 120 grams of solid NaOH;
- add this mass of NaOH into a 500.0-mL Erlenmeyer flask;
- fill approximately half of the flask with distilled water and stir gently to make sure that NaOH dissolves, if it doesn't, add more water and repeat the process;
- when NaOH fully dissolves, fill the flask to the mark.
Our solution is prepared.
Answer:
The images for the 3 sub-questions have been presented in the first, second and third attached image to this solution.
Explanation:
The reaction for the reaction between Hydrogen gas and Iodine to give Hydrogen iodide
H₂ (g) + I₂ (g) → 2 H I(g)
ΔH∘rxn = −9.5 kJ/mol r x n
a) From the information provided, the heat of reaction being negative shows that the reaction is an exothermic reaction with the products for an exothermic reaction having a lower energy content/level as the reactants for this reaction.
This shows that the energy diagram will have the products at a lower level than fe reactants.
The sketch of the energy diagram for this exothermic reaction is presented in the first attached image to this solution.
The heat of reaction is shown on the energy diagram, together with the activation energy, Ea, the heat content/energy level for the reactants (Hr) & products (Hr) and finally the ΔH∘rxn.
ΔH∘rxn = Hp - Hr
b) When a catalyst that speed up a chemical reaction is introduced to the chemical reaction, the catalyst goes about making the reaction faster by lowering the activation energy of the reaction.
The activation energy is the minimum energy that the reactants must possess to the able to form products. On the graph, it is denoted as Ea, the difference in energy between the peak of the curve and the emergy level of the reactants.
So, a catalyst will cause this activation energy to be lowered to a new level of Ea₁.
The energy diagram of this is presented in the second attached image to this answer.
The ΔH∘rxn remains unchanged for this introduction of catalyst.
The broken lines are used to represent the lowered activation energy action of the catalyst.
c) The reverse reaction will have opposite the energy properties of the initial reaction.
The initially exothermic reaction with higher heat content of reactants than products is now an endothermic reaction with heat content/energy level of the products higher than that of the reactants.
The energy diagram is presented in the third attached image.
The activation energy is bigger and the ΔH∘rxn is now positive and equal to 9.5 kJ/mol, the direct additive inverse of the ΔH∘rxn for the initial exothermic reaction.
Hope this Helps!!!
