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OpenStudy (anonymous):

how many grams of (N2H5)2C3H4O4 are in 1L of a 2m solution

OpenStudy (anonymous):

M(substance)=3Ar(C)+4Ar(N)+4Ar(O)+14Ar(H)=3x12+4x14+4x16+14x1=170 g/mol c=m/(MV) i.e. m=c.M.V=2x170x1=340 g

OpenStudy (anonymous):

muradi ty very much man i got the same answear i just have an issue witht he way the questions are worded if u can help me on a few more i would really owe u antyhing i can do please let me know

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