The mass of the aqueous solution in grams is obtained by subtracting mass of the 10.0 mL graduated cylinder alone from the mass of the cylinder + solution. Therefore mass of the solution is 2.951 g.
We have from the question, the following information;
Mass of empty 10.0 mL graduated cylinder to be 23.099 g
Mass of 10.0 mL graduated cylinder + 3.09 mL of solution to be 26.050g
Mass of aqueous solution in grams = mass of the cylinder + solution - mass of the 10.0 mL graduated cylinder alone
= 26.050g - 23.099 g
= 2.951 g
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Answer:
⇒ 3.312 E-13 mol Ni(OH)2 are soluble in 1 L of solution (NaOH)
Explanation:
S S 2S + 0.0218
∴ pKsp Ni(OH)2 = 15.8 = -Log Ksp
⇒ Ksp = 1.585 E-16 = [ Ni2+ ] * [ OH- ]²
∴ pH = 12.34
⇒ pOH = 14- 12.34 = 1.66 = - Log [ OH- ]
⇒ [ OH- ] = 0.0218 M
⇒ Ksp = 1.585 E-16 = S * ( 2S + 0.0218 )²
if we compare the concentration ( 0.0218 ) with the Ksp ( 1.585 E-16 ), then we can neglect the solubility as adding
⇒ 1.585 E-16 = S * ( 0.0218 )²
⇒ 1.585 E-16 = 4.786 E-4 * S
⇒ S = 3.312 E-13 mol/L
∴ soluble moles Ni(OH)2 = 3.312 E-13 mol/L * 1 L Sln = 3.312 E-13 moles
⇒ % S = ( 3.312 E-13 / 0.0218 ) * 100 = 1.513 E-11 %.....we can accept the previous assumption.
Answer:
what is late ? there is no attachment ?
