Answer:
-75 cm^3/min
Explanation:
Given from Boyle's law;
PV=C
From product rule;
VdP/dt + PdV/dt = dC/dt
but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min
PdV/dt = dC/dt - VdP/dt
dV/dt = dC/dt - VdP/dt/ P
substituting values;
dV/dt = 0 - (500 * 30)/200
dV/dt = -75 cm^3/min
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:
I don't really know what that is so here is a picture of it
Explanation:
Hey there :)
<em>Q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em> </em><em>How many km are in 5.6mm? </em>
<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>3 </em>
<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-6 </em>
<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-3 </em>
<em>=</em><em>></em><em> </em><em>5.6x10</em><em>^</em><em>6</em>
<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em>-</em>

<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em>-</em>
By using the formula-

As 1 with 6 zeros, we convert it into exponential form.

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

Now combine with given question.

Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ