Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{KOH}=0,1.1=0,1\left(mol\right)\)
PTHH: 2KOH + H2SO4 → K2SO4 + 2H2O
Mol: 0,1 0,1 0,1
b) \(V_{ddH_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(l\right)\)
c) \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,1+0,2}=0,333M\)
a) 2NaOH + H2SO4→ Na2SO4 + 2H2O
b) nNaOH = CMNaOH . V= 1. 0,1= 0,1mol
PTHH:
2NaOH + H2SO4 → Na2SO4 + 2H2O
2 1 1 2
0,1 0.05 0,05 0,1
VH2SO4 = 0,05/0,5 =0,1l
c) Vdd sau phản ứng = 0,1+0,1=0,2l
CM = 0,05/0,2 = 0,25M
![](https://rs.olm.vn/images/avt/0.png?1311)
nHCl=0,3.2=0,6(mol)
a) PTHH: CuO +2 HCl -> CuCl2 + H2O
0,3_______________0,6___0,3(mol)
b) mCuO=0,3.80=24(g)
c) VddCuCl2=VddHCl=0,3(l)
=>CMddCuCl2=0,3/0,3=1(M)
d) m(muối)=0,3.135=40,5(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PTHH: HCl + KOH → KCl + H2O
Mol: 0,4 0,4 0,4
b, \(V_{ddKOH}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\approx0,267\left(l\right)\)
c, \(C_{M_{ddKOH}}=\dfrac{0,4}{0,2+\dfrac{4}{15}}=\dfrac{6}{7}M\approx0,857M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PTHH: HCl + KOH → KCl + H2O
Mol: 0,4 0,4 0,4
b) \(V_{ddKOH}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\approx0,267\left(l\right)\)
c) \(C_{M_{ddKCl}}=\dfrac{0,4}{0,2+\dfrac{4}{15}}=\dfrac{6}{7}M\approx0,857M\)
\(a,PTHH:KOH+HCl\rightarrow KCl+H_2O\\ b,n_{KOH}=n_{HCl}=2\cdot0,1=0,2\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1\left(l\right)\\ c,C_{M_{KCl}}=\dfrac{0,2}{0,1+0,1}=1M\)
Đổi 100ml = 0,1 lít
Ta có: \(n_{KOH}=2.0,1=0,2\left(mol\right)\)
a. PTHH: \(KOH+HCl--->KCl+H_2O\)
b. Theo PT: \(n_{HCl}=n_{KOH}=0,2\left(mol\right)\)
\(\Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1\left(lít\right)=100\left(ml\right)\)
c. Ta có: \(V_{dd_{KCl}}=V_{dd_{HCl}}=0,1\left(lít\right)\)
Theo PT: \(n_{KCl}=n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{KCl}}=\dfrac{0,2}{0,1}=2M\)