Hóa học hòa tan 29g nacl vào 331g nacl2% tính c% dd thu được 10/10/2021 By Caroline hòa tan 29g nacl vào 331g nacl2% tính c% dd thu được
$mct_{NaCl(2)}$=$\frac{331.2}{100}$=6,62 (g) $mct_{NaCl(tổng)}$=$mct_{NaCl(1)}$+$mct_{NaCl(2)}$=29+6,62=35,62 (g)$mdd_{NaCl(sfu)}$=$mct_{NaCl(1)}$+$mdd_{NaCl(tfu)}$=29+331=360 (g) $Cphầntrăm_{NaCl(sfu)}$=$\frac{35,62.100}{360}$≈9,8944% Trả lời
Đáp án: VixionD mctNaCl(2)mctNaCl(2)=331.2100331.2100=6,62 (g) mctNaCl(tổng)mctNaCl(tổng)=mctNaCl(1)mctNaCl(1)+mctNaCl(2)mctNaCl(2)=29+6,62=35,62 (g)mddNaCl(sfu)mddNaCl(sfu)=mctNaCl(1)mctNaCl(1)+mddNaCl(tfu)mddNaCl(tfu)=29+331=360 (g) CphầntrămNaCl(sfu)CphầntrămNaCl(sfu)=35,62.10036035,62.100360≈9,8944% Trả lời
$mct_{NaCl(2)}$=$\frac{331.2}{100}$=6,62 (g)
$mct_{NaCl(tổng)}$=$mct_{NaCl(1)}$+$mct_{NaCl(2)}$=29+6,62=35,62 (g)
$mdd_{NaCl(sfu)}$=$mct_{NaCl(1)}$+$mdd_{NaCl(tfu)}$=29+331=360 (g)
$Cphầntrăm_{NaCl(sfu)}$=$\frac{35,62.100}{360}$≈9,8944%
Đáp án:
mctNaCl(2)mctNaCl(2)=331.2100331.2100=6,62 (g)
mctNaCl(tổng)mctNaCl(tổng)=mctNaCl(1)mctNaCl(1)+mctNaCl(2)mctNaCl(2)=29+6,62=35,62 (g)
mddNaCl(sfu)mddNaCl(sfu)=mctNaCl(1)mctNaCl(1)+mddNaCl(tfu)mddNaCl(tfu)=29+331=360 (g)
CphầntrămNaCl(sfu)CphầntrămNaCl(sfu)=35,62.10036035,62.100360≈9,8944%