Explanation:
Reaction equation showing alpha decay in Uranium-238 is as follows.

It is known that an alpha particle is basically a helium nucleus and it contains 2 protons and 2 neutrons.
Symbol of an alpha particle is
.
As atomic mass or weight is the sum of total number of protons and neutrons present in an atom.
Hence, the atomic weight of the alpha particle is (2 + 2) = 4.
Answer:
It is A).
Explanation:
Silver (Ag) goes from the pure metal to Ag+ losing 1 electron so it is oxidised.
The hydrogen ion gains electrons and is reduced.
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17%
concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride
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.