No its a example of heterogeneous mixture
Answer:
matter cannot be created nor destroyed
Explanation:
Answer:
Kc = [CO2], that is to say it is equal to the concentration of CO2
Explanation:
It is a heterogeneous equilibrium since the substances that participate in the reaction are in different phases
In the heterogeneous limestone decomposition reaction:
CaCO3(s) --> CaO(s) + CO2(g)
The equilibrium constants are:
Kc = [CO2(g)]; Kp = PCO2(g); Kc = Kp (R T)^
−(1−0) = Kp (R T)^
−1
The equilibrium situation is not affected by the amount of solid or liquid, as long as these substances are present.
The equilibrium constant is independent of the amounts of solids and liquids in equilibrium.
<span>Moles = 0.252
Molarity = 1.07
This question is badly worded. You're asking for moles and I suspect you really want molarity. The number of moles of ammonium chloride you have in the solution will remain constant regardless of the volume of the solution. However, the molarity of the solution will differ depending upon how concentrated it is. So I'll give you both the number of moles of ammonium chloride you have, and the molarity of the resulting solution. Please talk to your teacher if you're confused by the difference between moles and molarity.
The formula for ammonium chloride is NH4Cl. So let's calculate it's molar mass. Start by looking up the associated atomic weights.
Atomic weight nitrogen = 14.0067
Atomic weight hydrogen = 1.00794
Atomic weight chlorine = 35.453
Molar mass NH4Cl = 14.0067 + 4 * 1.00794 + 35.453 = 53.49146 g/mol
Moles NH4Cl = 13.5 g / 53.49146 g/mol = 0.252376735 mol
Molarity is defined as moles per liter, so let's divide the number of moles we have by the volume in liters. So:
0.252376735 mol / 0.235 l = 1.073943551 M
Rounding to 3 significant figures gives: 0.252 moles, 1.07 molarity.</span>
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density 
; we have:
The average atomic weight is:
So; in terms of vanadium, the Concentration of iron is:
From a unit cell volume 
where;
= number of Avogadro constant.
SO; replacing with 
; with 
; 
with 
and
with 
Then:
![a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2FA_%7BFe%7D%20%5D%20%2B%20%5BC_v%2FA_v%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2F%5Crho_%7BFe%7D%20%5D%20%2B%20%5BC_v%2F%5Crho_v%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100-C_v%29A_%7Bv%7D%20%5D%20%2B%20%5BC_v%2FA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100-C_v%29%2F%5Crho_%7Bv%7D%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100A_%7Bv%7D-C_vA_%7Bv%7D%29%20%5D%20%2B%20%5BC_vA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100%5Crho_%7Bv%7D%20-%20C_v%20%5Crho_%7Bv%7D%29%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
Replacing the values; we have:
