Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
And the undergoing chemical reaction:
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:
Next, the moles of magnesium chloride consumed by the sodium fluoride:
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
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:
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.
Answer:
see note under explanation
Explanation:
When describing system and surroundings the system is typically defined as the 'object of interest' being studied and surroundings 'everything else'. In thermodynamics heat flow is typically defined as endothermic or exothermic. However, one should realize that the terms endothermic and exothermic are in reference to the 'system' or object of interest being studied. For example if heat is transferred from a warm object to a cooler object it is imperative that the system be defined 1st. So, with that, assume the system is a warm metal cylinder being added into cooler water. When describing heat flow then the process is exothermic with respect to the metal cylinder (the system) but endothermic to the water and surroundings (everything else).
Answer:
Answer: CH₃ and C₂H₆ have same empirical formula.
Explanation:
it just compares in that it its the same
Answer:
The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.
Explanation:
In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.
H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.
All the answers are on the actual periodic table. You should never be told to remember it so I think this is a recourse you are allowed to look at whilst doing your homework lol :) all the answers are written on it hope it helps
