It will depend on the chemical, if it's a chemical compound, yes it can be decomposed chemically into simpler substances, further elements that constitute or make up the given compound.
Mixtures are separated by physical means as there are still separate entities as the substances required to make the mixture do not chemically combine or become altered chemically.
Elements cannot be decomposed into simpler substances as the simplest substance of matter is an element itself. This cannot be decomposed by chemical means.
The total number of atoms remains unchanged, <em>always.</em> But the chemical properties change :)
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Answer:
Solubility, length, hardness, color, mass, density, weight, volume, boiling, and point.
Explanation: Hope this helps
Here is the correct question
You mix 125 mL of 0.170 M CsOH with 50.0 mL of 0.425 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 20.20 °C before mixing to 22.17 °C after the reaction. What is the enthalpy of reaction per mole of ? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J/g · K. Enthalpy of reaction = kJ/mol
Answer:
75.059 kJ/mol
Explanation:
The formula for calculating density is:
Making mass the subject of the formula; we have :
mass = density × volume
which can be rewritten as:
mass of the solution = density × volume of the solution
= 1.00 g/mL × (125+ 50 ) mL
= 175 g
Specific heat capacity = 4.2 J/g.K
∴ the energy absorbed is = mcΔT
= 175 × 4.2 × (22.17 - 20.00) ° C
= 1594.95 J
= 1.595 J
number of moles of CsOH = 
= 0.2125 mole
Therefore; the enthalpy of the reaction = 
= 
= 75.059 kJ/mol
The reaction is:
2 NO₂ (g) + F₂ (g) ⇆ 2 NO₂F (g)
The stoichiometric coefficients of the substances balance out each other to obey the Law of Definite Proportions. Now, you have to note that determining the reaction rate expression is specific to a certain type of reaction. So, this are determined empirically through doing experiments. But in chemical reaction engineering, to make things simple, you assume that the reaction is elementary. This means that the order of a reaction with respect to a certain substance follows their individual stoichiometric coefficients. What I'm saying is, the stoichiometric coefficients are the basis of our reaction rate orders. For this reaction, the rate order is 2 for NO₂, 1 for F₂ and 2 for NO₂F. When the forward and reverse reactions are in equilibrium, then it applies that:
Reaction rate of disappearance of reactants = Reaction rate of formation of products.
Therefore, we can have two reaction rate constants for this. But since the conditions manipulated are the reactant side, let's find the expression for reaction rate of disappearance of reactants.
-r = k[NO₂]²[F₂]
The negative sign before r signifies the rate of disappearance. If it were in terms of the product, that would have been positive. The term k denotes for the reaction rate constant. That is also empirical. As you can notice the stoichiometric coefficients are exponents of the concentrations of the reactants. Let's say initially, there are 1 M of NO₂ and 1 M of F₂. Then,
-r = k(1)²(1)
-r = k
Now, if we change 1 M of NO₂ by increasing it to its half, it would now be 1.5 M NO₂. Then, if we quadruple the concentration of F₂, that would be 4 M F₂. Substituting the values:
-r = k(1.5)²(4)
-r = 9k
So, as you can see the reaction rate increase by a factor of 9.
