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Các bạn giúp mình hoàn thiện 2 bảng naày với. Nhanh hộ minhf nha, mình đang cần gấp

Nguyễn Hoàng Minh · Accepted Answer

$NaCl$$Ca\left(OH\right)_2$$BaCl_2$$KOH$$CuSO_4$$m_{CT}$$30g$$0,148g$$\dfrac{150\cdot20\%}{100\%}=30\left(g\right)$$42g$$3g$$m_{H_2O}$$170g$$199,852g$$120g$$270g$$17g$$m_{dd}$$200g$$\dfrac{0,148\cdot100\%}{0,074\%}=200\left(g\right)$$150g$$312g$$\dfrac{3\cdot100\%}{15\%}=20\left(g\right)$$C\%$$\dfrac{30}{200}\cdot100\%=15\%$$0,074\%$$20\%$$\dfrac{42}{312}\cdot100\%\approx13,46\%$$15\%$ Câu trả lời:  $NaCl$$Ca\left(OH\right)_2$$BaCl_2$$KOH$$CuSO_4$$m_{CT}$$30g$$0,148g$$\dfrac{150\cdot20\%}{100\%}=30\left(g\right)$$42g$$3g$$m_{H_2O}$$170g$$199,852g$$120g$$270g$$17g$$m_{dd}$$200g$$\dfrac{0,148\cdot100\%}{0,074\%}=200\left(g\right)$$150g$$312g$$\dfrac{3\cdot100\%}{15\%}=20\left(g\right)$$C\%$$\dfrac{30}{200}\cdot100\%=15\%$$0,074\%$$20\%$$\dfrac{42}{312}\cdot100\%\approx13,46\%$$15\%$

Nguyễn Hoàng Minh · Answer

$A$$B$Trả lời1/...$a,10\%$$n_{HCl}=0,5\cdot2=1\left(mol\right)\Rightarrow B$2/...(sửa đề là $m_{CT}$)$b,1mol$$m_{CT_{H_2SO_4}}=\dfrac{250\cdot20\%}{100\%}=50\left(g\right)\Rightarrow C$3/...$c,50g$$V_{dd_{NaOH}}=\dfrac{0,2}{1}=0,2\left(l\right)\Rightarrow D$4/...$d,0,2\text{ lít}$$C\%=\dfrac{25}{250}\cdot100\%=10\%\Rightarrow A$ $e,500ml$  Câu trả lời: $A$$B$Trả lời1/...$a,10\%$$n_{HCl}=0,5\cdot2=1\left(mol\right)\Rightarrow B$2/...(sửa đề là $m_{CT}$)$b,1mol$$m_{CT_{H_2SO_4}}=\dfrac{250\cdot20\%}{100\%}=50\left(g\right)\Rightarrow C$3/...$c,50g$$V_{dd_{NaOH}}=\dfrac{0,2}{1}=0,2\left(l\right)\Rightarrow D$4/...$d,0,2\text{ lít}$$C\%=\dfrac{25}{250}\cdot100\%=10\%\Rightarrow A$ $e,500ml$

	\(NaCl\)	\(Ca\left(OH\right)_2\)	\(BaCl_2\)	\(KOH\)	\(CuSO_4\)
\(m_{CT}\)	\(30g\)	\(0,148g\)	\(\dfrac{150\cdot20\%}{100\%}=30\left(g\right)\)	\(42g\)	\(3g\)
\(m_{H_2O}\)	\(170g\)	\(199,852g\)	\(120g\)	\(270g\)	\(17g\)
\(m_{dd}\)	\(200g\)	\(\dfrac{0,148\cdot100\%}{0,074\%}=200\left(g\right)\)	\(150g\)	\(312g\)	\(\dfrac{3\cdot100\%}{15\%}=20\left(g\right)\)
\(C\%\)	\(\dfrac{30}{200}\cdot100\%=15\%\)	\(0,074\%\)	\(20\%\)	\(\dfrac{42}{312}\cdot100\%\approx13,46\%\)	\(15\%\)

\(A\)	\(B\)	Trả lời
1/...	\(a,10\%\)	\(n_{HCl}=0,5\cdot2=1\left(mol\right)\Rightarrow B\)
2/...(sửa đề là \(m_{CT}\))	\(b,1mol\)	\(m_{CT_{H_2SO_4}}=\dfrac{250\cdot20\%}{100\%}=50\left(g\right)\Rightarrow C\)
3/...	\(c,50g\)	\(V_{dd_{NaOH}}=\dfrac{0,2}{1}=0,2\left(l\right)\Rightarrow D\)
4/...	\(d,0,2\text{ lít}\)	\(C\%=\dfrac{25}{250}\cdot100\%=10\%\Rightarrow A\)
	\(e,500ml\)

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