We can calculate for temperature by assuming the equation
for ideal gas law:
P V = n R T
Where,
P = pressure = 1.80 atm
V = volume = 18.2 L
n = number of moles = 1.20 moles
R = gas constant = 0.08205746 L atm / mol K
Substituting to the given equation:
T = P V / n R
T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm /
mol K)
T = 332.70 K
We can convert K unit to ˚C unit by subtracting 273.15
to Kelvin, therefore
T = 59.55 ˚<span>C</span>
Electronegativity<span> is the measure of the ability of an atom to attract electrons to itself. Fluorine is the most </span>electronegative<span> element and francium is one of the least</span>electronegative<span>. ... The </span>molecule's polarity<span> will be determined on the negative and positive regions on the outer atoms in the </span>molecule<span>.</span>
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:
I am sure that the C one is correct