How many milliliters of a 1: 5000 w/v solution of the preservative benzalkonium chloride can be made from 125 ml of a 0.2% w/v solution of the preservative? a. 1250 ml b. 2500 ml c. 625 ml d. 1875 ml

a. 1250 mL

Explanation:

We use the C1V1=C2V2 equation to solve this problem, with:

C1=1/5000=0,0002 w/v,             V1=?

C2=0,2%=0,2/100=0,002 w/v     V2=125 mL

Then:

V1=C2V2/C1=0,002*125/0,0002=125O mL

Thus, 1250 mL of the preservative 1:5000 w/v can be made from 125 mL of a 0,2% w/v solution.

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

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