Answer : The mass of produced is, 7.74 grams
Solution : Given,
Mass of = 13.3 g
Molar mass of = 227 g/mole
Molar mass of = 44 g/mole
First we have to calculate the moles of .
Now we have to calculate the moles of
The balanced chemical reaction is,
From the reaction, we conclude that
As, 4 mole of react to give 12 mole of
So, 0.0586 moles of react to give moles of
Now we have to calculate the mass of
Thus, the mass of produced is, 7.74 grams
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The physical state of water at its boiling point temperature of 100 degree Celsius will be both liquid state as well as gaseous state. This is because at its boiling point of 100 degree celsius the liquid state of water starts changing into its Gaseous state (steam).
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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