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:
p-fluoronitrobenzene and sodium phenoxide is more appropriate
Explanation:
An ipso substitution is required to form p-nitrophenyl phenyl ether.
For this ipso substitution, an alkoxide anion needs to attack as a nucleophile at the carbon atom attached to fluorine atom and thereby substitute that F atom.
p-nitrophenoxide is an weak nucleophile as compared to phenoxide due to presence of electron withdrawing resonating effect of nitro group at para position.
p-fluoronitrobenzene is a good choice for nucleophilic attack by alkoxide anion as compared to fluorobenzene due to higher positive charge density at carbon atom directly attached to F atom. Higher positive charge density arises due to presence of electron withdrawing resonating effect og nitro group at para position.
So, p-fluoronitrobenzene and sodium phenoxide is more appropriate
Answer:
CH3OH and NADH
Explanation:
The given chemical reaction is an redox reaction in which reduction and oxidation take place.
In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.
In the given redox reaction: CH3OH + NAD --> CH2O + NADH
NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal) by losing electrons.
So, CH3OH (methanol) and NADH are the reduced forms while NAD and CH2O (methanal) are oxidized forms.
