Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2.
mmuối trong dd 8%=190.\(\dfrac{8}{100}\)=15,2(g)
C% dd muối=\(\dfrac{15,2+10}{10+190}.100\%=12,6\%\)
Elly Phạm giải hộ nó bài đầu cái,tui ko muốn làm nữa
\(m_{d^2}=10+100=110\left(g\right)\\ m_{ct\text{ }100\left(g\right)\text{ }d^2\text{ }8\%}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{100\cdot8}{100}=8\left(g\right)\\ m_{ct\text{ }trong\text{ }110\left(g\right)\text{ }d^2pha\text{ }pha\text{ }ch\text{ế}}=8+10=18\left(g\right)\\ \Rightarrow C\%=\dfrac{m_{ct}}{m_{d^2}}\cdot100=\dfrac{18}{110}\cdot100=16,36\%\)
Câu 1:
\(a\text{) }P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{6\sqrt{x}-2}{2x-\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{3x-3\sqrt{x}}\\ ĐKXĐ:x>0;x\ne1\\ \Rightarrow P=\left(\dfrac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{3\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}+2+3\sqrt{x}-3-6\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1}{3\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\\ =\dfrac{3\sqrt{x}}{2\sqrt{x}+1}\)
\(\text{b) Ta có }:x=\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\\ =2\left(\dfrac{\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\dfrac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right)\\ =2\cdot\dfrac{\sqrt{3}+1-\sqrt{3}+1}{3-1}=2\)
Với \(x>0;x\ne1\)
\(Khi\text{ }x=2\left(T/m\right)\Rightarrow P=\dfrac{3\sqrt{x}}{2\sqrt{x}+1}=\dfrac{3\sqrt{2}}{2\sqrt{2}+1}\\ =\dfrac{3\sqrt{2}\left(2\sqrt{2}-1\right)}{\left(2\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)}\\ =\dfrac{12-3\sqrt{2}}{8-1}=\dfrac{12-3\sqrt{2}}{7}\)
Ta có: \(C\%=20\%=\dfrac{m_{NaCl}}{250}.100\%\)
\(\Rightarrow m_{NaCl}=50\left(g\right)\)
\(m_{\text{dd}}=\dfrac{100.0,5}{2,5}=20g\\ m_{H_2O}=20-0,5=19,5g\)
\(V_{NaClsau}=3+4=7\left(l\right)\)
\(n_{NaCl0,5M}=0,5.3=1,5\left(mol\right)\)
\(n_{NaCl1,5M}=1,5.4=6\left(mol\right)\)
\(C_{M_{NaClsau}}=\dfrac{n}{V}=\dfrac{1,5+6}{7}=\dfrac{15}{14}M\)
\(a.\)
\(m_{dd}=10+40=50\left(g\right)\)
\(C\%=\dfrac{10}{50}\cdot100\%=20\%\)
\(b.\)
\(m_{KOH}=0.25\cdot56=14\left(g\right)\)
\(m_{dd_{KOH}}=14+36=50\left(g\right)\)
\(C\%_{KOH}=\dfrac{14}{50}\cdot100\%=28\%\)
Câu 1:
Trong 500g dd muối ăn, lượng muối ăn chiếm:
\(m_{NaCl}=\dfrac{6.500}{100}=30\left(g\right)\)
Câu 4:
Khối lượng muối ăn có trong 140g dd NaCl 10%:
\(m_{NaCl}=\dfrac{140.10}{100}=14\left(g\right)\)
=> \(m_{H_2O}=140-14=126\left(g\right)\)
Vậy: để có 140g dd NaCl 10% ta cần cho 14g NaCl vào 126g nước.
2) Ah Ddd=1,1g/ml moi dung , hinh nhu ghi sai de
a, Theo de bai ta co
So gam muoi an co trong 50d dd muoi an 8%la
mct=\(\dfrac{mdd.C\%}{100}=\dfrac{50.8\%}{100\%}=4g\)
So gam muoi an co trong 400g dd muoi an 20% la
mct=\(\dfrac{mdd.C\%}{100\%}=\dfrac{400.20\%}{100\%}=80g\)
So gam muoi an co trong dd muoi sau khi tron la
mct= 4+ 80 =84 g
So gam dd muoi sau khi tron la
mdd=50 + 400=450 g
\(\Rightarrow\) C%=\(\dfrac{mct}{mdd}.100\%=\dfrac{84}{450}.100\%\approx18,7\%\)
b, Theo cau a, ta co
khoi luong dd muoi sau khi tron la
mdd =50+400=450 g
\(\Rightarrow\) The tich dung dich muoi sau khi tron la
Vdd=\(\dfrac{mdd}{Ddd}=\dfrac{450}{1,1}\approx409,1ml\)
Ad:C%=\(\frac{m_g}{m_{dd}}\)=8%→mg(muối trg dd)=15,2g →C2%=\(\frac{15,2+10}{190}\)\(\times\)100=13,2%