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
For an approximate result, multiply the volume value by 3.785
Answer ≈ 56.7812
After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
More about half-life: brainly.com/question/1160651
#SPJ4
C is not a property of metals