Question

Cho mik hỏi câu f 4 nha MN😘😘😘

Answer 1

\(n_{C_{12}H_{22}O_{11}\left(1\right)}=3\cdot1=3\left(mol\right)\)

\(n_{C_{12}H_{22}O_{11}\left(2\right)}=3\cdot2=6\left(mol\right)\)

\(C_{M_{C_{12}H_{22}O_{11}}}=\dfrac{3+6}{3+3}=1.5\left(M\right)\)

Answer 2

Ta có: $n_{duong1}=3(mol);n_{duong2}=6(mol)$

Do đó $C_{M/duongsau}=\frac{3+6}{3+3}=1,5M$

Answer 3

V dd = 3 + 3 = 6(lít)

n đường = 3.1 + 3.2 = 9(mol)

=> CM đường = 9/6 = 1,5M