4 years ago

How many moles of koh are required to prepare 0.80 l of 6.4 m koh? 0.13 mol koh 5.1 mol koh 7.2 mol koh 8.0 mol koh?

1 answer:

wel4 years ago

8 0

Answer is: 5,1 mol KOH.
V (KOH) = 0,8l = 0,8 dm³.
c (KOH) = 6,4 mol/dm³.
V - volume of solution.
c - concentration of solution.
c = n÷V
n - amount of substance.
n (KOH) = c·V = 6,4 mol/dm³ · 0,8 dm³ = 5,12 mol.
Potassium hydroxide (KOH) can<span> </span>dissolve<span> in water at room temperature.</span>

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$m_{Ag_2S}^{theoretical}=4.27gAgNO_3*\frac{1molAgNO_3}{169.87gAgNO_3} *\frac{1molAg_2S}{2molAgNO_3}*\frac{247.8gAg_2S}{1molAg_2S}\\\\m_{Ag_2S}^{theoretical}=3.11gAg_2S$

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