Answer:
<u>25 to 1 fuel mix 1 gallon</u>
Explanation:
Add two of the 2.5 ounce containers to 1 gallon of gas. For a 50:1 ratio add 1 container 2.5 ounces to 1 gallon of gas. 50=parts of fuel 1=parts of oil . For 25:1 and 2 of the 2.5 oz containers to 1 gallon of gas.
Answer:
well the world would be shocked
Explanation:
and suprise
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:
53.85%
Explanation:
Data obtained from the question include:
Mass of antimony (Sb) = 27.6g
Mass of Fluorine (F) = 32.2g
Mass of compound = 59.8g
Percentage composition of fluorine (F) =..?
The percentage composition of fluorine can be obtained as follow:
Percentage composition of fluorine = mass of fluorine/mass of compound x 100
Percentage composition of fluorine = 32.2/59.8 x 100
= 53.85%
Therefore, the percentage composition of fluorine in the compound is 53.85%