Answer:
hi
Explanation:
this is my first time on this app so what do i do
"A red giant is a star that has exhausted the primary supply of hydrogen fuel at its core. An average-sized star like our Sun will spend the final 10 percent of its life as a red giant. In this phase, a star's surface temperature drops to between 3,140 and 6,741°F (1,727 and 3,727°C) and its diameter expands to 10 to 1,000 times that of the Sun. The star takes on a reddish color, which is what gives it its name."
Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol
Moles=14.5g / 58.14g/mol
=0.249
Therefore there are approx 0.249 moles in a 14.5g sample of C4H10
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