Answer:

Explanation:
Hello!
In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:
![[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D1.672%5Cfrac%7BmolMgCl_2%7D%7BL%7D%2A%5Cfrac%7B95.211gMgCl_2%7D%7B1molMgCl_2%7D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D)
Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:
![[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D%2A%5Cfrac%7B1L%7D%7B1000mL%7D%2A%5Cfrac%7B1mL%7D%7B1.137g%7D%3D0.14)
Which is also the by-mass fraction and in percent it turns out:

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Answer:
The reaction will move to the left.
Explanation:
<em>Ba(OH)₂ = Ba²⁺ + 2OH⁻,</em>
<em>Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.</em>
- If H⁺ ions are added to the equilibrium:
H⁺ will combine with OH⁻ to form water.
<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>
<em />
<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.
- The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.
<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>
Answer:
higher, higher
Explanation:
It takes more energy to rip apart stronger bonds (that's mostly just common sense there). The boiling point increases because it would take more energy to get the molecules to go from a stuck together liquid, to separating in a gaseous form.