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1,825 phút bạn nhé

109,5 giờ = 109 giờ 30 phút

109,5 giờ = ... giờ ... phút

TL: 109,5 giờ = 109 giờ 30 phút

học tốt ~~~

109,5-67,23-9,5

=(109,5-9,5)-67,23

=100-67,23

=32,77

= (109,5-9,5)-67,23

\(a,\sqrt{52}.\sqrt{13}=\sqrt{2^2.13}.\sqrt{13}\)

\(=2\sqrt{13}.\sqrt{13}=2.13=26\)

\(b,\sqrt{146,5^2-109,5^2}+27.256\)

\(=\sqrt{\left(146,6-109,5\right)\left(146,5+109,5\right)}+27.256\)

\(=\sqrt{37.256}+27.256\)

\(=\sqrt{4^2.37}+27.16=4\sqrt{37}+4.4.27\)

\(=4\left(\sqrt{37}+108\right)\)

nZn = 13:65 = 0,2 mol
nHCl = 10%:100% x 109,5 : 36,5=0.3 mol
PTHH 
Zn + 2HCl --> ZnCl2 + H2 
Theo pt
0,2 / 1 > 0,3 / 2 nên Zn dư, phương trình tính theo HCl
nZnCl2 = 1/2nHCl = 0,15 mol
=) mZnCl2 = 0,15.136=20.4 g

\(n_{ZnO}=\dfrac{1,62}{81}=0,02\left(mol\right)\)

\(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂O

Mol:     0,02      0,3        0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{0,3}{1}\) ⇒ ZnO hết, HCl dư

m_{dd sau pứ} = 1,62+109,5 = 111,12 (g)

\(C\%_{ddZnCl_2}=\dfrac{0,02.136.100\%}{111,12}=2,45\%\)

\(C\%_{ddHCl}=\dfrac{\left(0,3-0,04\right).36,5.100\%}{111,12}=8,54\%\)

\(n_{ZnO}=\dfrac{1,62}{81}=0,02\left(mol\right)\)

\(m_{HCl}=109,5.10\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: ZnO + 2HCl → ZnCl₂ + H₂

Mol:     0,02    0,04        0,02      0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{0,3}{2}\) ⇒ ZnO hết, HCl dư

m_{dd sau pứ} = 1,62+109,5-0,02.2 = 111,08  (g)

\(C\%_{ZnCl_2}=\dfrac{0,02.136.100\%}{111,08}=2,45\%\)

\(C\%_{HCldư}=\dfrac{\left(0,3-0,04\right).36,5.100\%}{111,08}=8,54\%\)