Sarah's class is studying soils. Her teacher, Ms. Jones, stated that plants will growth in any soil type. Students in Sarah's cl
2 answers:
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Answer:
P=atm
Explanation:
The problem give you the Van Der Waals equation:
First we are going to solve for P:
Then you should know all the units of each term of the equation, that is:
where atm=atmosphere, L=litters, K=kelvin
Now, you should replace the units in the equation for each value:
Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:
Then operate the fraction subtraction:
P=
And finally you can find the answer:
P=atm
Now solving for b:
Replacing units:
Multiplying and dividing units,(please see the second photo below), we have:
Answer:
Explanation:
Iron is Fe, nitrogen is N. Nitrogen is diatomic, which means it occurs as a molecular pair by itself. Iron III nitride has a chemical formula of FeN because nitrogen has a charge of 3-, and iron III tells us the iron has a charge of 3+ so you just need one of each to make the charges balance and the compound neutral.
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