Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
The matter wasn't destroyed because albert einsteins theory of the black hole says that matter cannot be destroyed if it is burned into the same color of a black hole. this eqation proves my thesis:
²²↑↑Ф
To solve this we assume
that the gas is an ideal gas. Then, we can use the ideal gas equation which is
expressed as PV = nRT. At a constant temperature and number of moles of the gas
the product of PV is equal to some constant. At another set of condition of
temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
P2 = P1 x V1 / V2
P2 = 2.0 x 1.5 / 3
<span>P2 = 1 atm</span>
Answer: 8moles
Explanation:
The reaction below shows the formation of 2 moles of water from 2 moles of hydrogen and 1 mole of oxygen respectively.
2H2(g) + O2 (g) --> 2H2O(l)
So, if 1 mole of O2 produce 2 mole of H2O
4 moles of O2 will produce Z mole of H2O
To get the value of Z, cross multiply
1 x Z = 4 x 2
Z = 8
So, the equation will be 8H2(g) + 4O2 (g) --> 8H2O(l)
Thus, 4 moles of O2 will produce 8moles of H2O .
Answer:
Explanation:
In theory, not much of anything. The vast majority of nitrates are water soluble. Aside, not sure what chemistry level you are at but you will probably be asked to know or memorize some solubility rules. This, for lack of a better phrase, Nitrate rule, is near spot on. With one exception—a rare one—all metal cationic nitrates are soluble in water. All of them. So, assuming you are talking about aqueous, water-based solutions of these salts and mixing them together, I expect nothing to occur. Both solutions, I believe are colorless in water and will thus remain so. If you had say a solution of Iron (III) nitrate and copper (II) nitrate, slightly different story. Both are colorful solutions and I would think you might see blending of colors but no reaction; no precipitate will form. You will probably learn about markers of a chemical reaction. One of these is a color change. Note, you should read this as a change of color from what you previously had. Going from red to blue or colorless to colored (or vice versa) is a strong indication of a reaction (e. g. evidence of bond-breaking and bond-formation). The mere mixing of colors does not constitute a chemical reaction.
