Answer : behavior in a field experiment is more likely to reflect real life because of its natural setting, i.e. higher ecological validity than a lab experiment
Explanation:
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
![MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-](https://tex.z-dn.net/?f=MgF_2%28s%29%5Crightleftharpoons%20Mg%5E%7B2%2B%7D%2B2F%5E-)
And the undergoing chemical reaction:
![MgCl_2+2NaF\rightarrow MgF_2+2NaCl](https://tex.z-dn.net/?f=MgCl_2%2B2NaF%5Crightarrow%20MgF_2%2B2NaCl)
We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:
![n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2](https://tex.z-dn.net/?f=n_%7BMgCl_2%7D%3D0.3L%2A1.1x10%5E%7B-3%7Dmol%2FL%3D3.3x10%5E%7B-4%7DmolMgCl_2)
Next, the moles of magnesium chloride consumed by the sodium fluoride:
![n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2](https://tex.z-dn.net/?f=n_%7BMgCl_2%7D%5E%7Bconsumed%7D%3D0.5L%2A1.2x10%5E%7B-3%7DmolNaF%2FL%2A%5Cfrac%7B1molCaCl_2%7D%7B2molNaF%7D%20%3D3x10%5E%7B-4%7DmolMgCl_2)
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
![n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2](https://tex.z-dn.net/?f=n_%7BMgF_2%7D%3D3x10%5E%7B-4%7DmolMgCl_2%2A%5Cfrac%7B1molMgF_2%7D%7B1molMgCl_2%7D%3D3x10%5E%7B-4%7DmolMgF_2)
Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
![[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M](https://tex.z-dn.net/?f=%5BMg%5E%7B2%2B%7D%5D%3D%5Cfrac%7B3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D3.75x10%5E%7B-4%7DM)
![[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M](https://tex.z-dn.net/?f=%5BF%5E-%5D%3D%5Cfrac%7B2%2A3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D7.5x10%5E%7B-4%7DM)
Thereby, the reaction quotient is:
![Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}](https://tex.z-dn.net/?f=Q%3D%283.75x10%5E%7B-4%7D%29%287.5x10%5E%7B-4%7D%29%5E2%3D2.11x10%5E%7B-10%7D)
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Answer:
In He2 molecule,
Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.
Molecular Orbitals thus formed are:€1s2€*1s2
It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .
Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2
Bond Order =Nb-Na/2
Bond Order =2-2/2=0
Since the bond order is zero so that He2 molecule does not exist.
Explanation:
Answer:
covalent
Explanation:
The carbon and the nitrogen very often form bonds in nature, carbon-nitrogen bonds, which are covalent types of bonds. In fact, the bonds between the carbon and nitrogen are one of the most abundant in the biochemistry and the organic chemistry. The bonds between these two can be double bonds, as well as triple bonds. The carbon-nitrogen bonds have the tendency to be strongly polarized toward the nitrogen.