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Mathematics 14 Online
OpenStudy (anonymous):

how many grams of K2SO4 molar mass=174 g/mol would be needed to prepare 4.00 L of a 0.0510 M solution?

OpenStudy (anonymous):

the answer is 35.5 i found out, but does anyone know how to get that

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i need like the work showed

OpenStudy (anonymous):

ok !!

OpenStudy (anonymous):

10.00 mL of potassium was diluted to make exactly 250.00 mL of solution, this should give you a starting point.

OpenStudy (anonymous):

you got that from chacha

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

im stuck on that.

OpenStudy (anonymous):

0.051*4*174 = 35.496g

OpenStudy (anonymous):

each litre should contain exactly 0.051 moles of of K2SO4 which should weigh 174*0.051

OpenStudy (anonymous):

thank you!!

OpenStudy (anonymous):

ok

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