The concentration of calcium carbonate does not appear in the rate law because it is a solid and its concentration is constant. The molar density, which is fixed, is the concentration of the solid. In this chemical reaction, only the amount of hydrochloric acid affects the rate of reaction.
a. Spectator ions are the ones that are present but does not enter into the actual chemical reaction.
b. Na⁺ and NO₃⁻ ions are spectator ions involved in this above mentioned reaction
<u>Explanation:</u>
- In redox reactions, for example reaction of aqueous silver nitrate and aqueous sodium chloride, silver chloride settles down as a precipitate by the addition of Ag⁺ ion from AgNO₃ and Cl⁻ ions from NaCl .
- The other 2 ions that is NO₃⁻ ion and Na⁺ ion in the reaction do not enter into the reaction and they are called as Spectator ions.
This can be expressed as the following equation,
AgNO₃(aq) + NaCl(aq) →AgCl(s) + NaNO₃
When you change the number of protons in an atom, you will change the atom from one element to a different element. Sometimes, when you add a proton to an element, the element will become radioactive. If you change the number of electrons in an atom, you will get an ion of the element.
Answer:
-54 kJ/mol
Explanation:
Given that:
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol
i.e
50 ml of 1.0 M HCl + 50 ml of 1.0 M NaOH -----> -54 kJ/mol
If he repeat the same experiment with :
100 ml of 1.0 M HCl + 100 ml of 1.0 M NaOH. ------> ????
From The experiment; the molar enthalpy of change of the acid-base neutralization reaction will be -54 kJ/mol
This is because : The second reaction requires 50 ml in order to neutralize the reaction, then the remaining 50 ml will be excess, Hence, there is no change in the enthalpy of the reaction.
Similarly; we can assume that :
In the first reaction; P moles of is used to liberate Q kJ heat ; then the change in molar enthalpy will be Q/P (kJ/mol).
SO; when he used 100 ml ;
then the amount of moles used is double, likewise the heat liberated will be doubled ;
So;
2P moles is used to liberate 2Q kJ heat ;
2P/2Q = Q/P ( kJ/mol) = -54 kJ/mol
Explanation:
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