D. CuCl2 copper(2)chloride
Most likely b,c or d sorry but idk I’m just helping other people answer by eliminating one answer
Answer:
1. 0.0154mole of PbS
2. Double displacement reaction
Explanation:
First, let write a balanced equation for the reaction. This is illustrated below:
Pb(CH3COO)2 + H2S —> PbS + 2 CH3COOH
Molar Mass of Pb(CH3COO)2 = 207 + 2(12 + 3 + 12 + 16 +16) = 207 + 2(59) = 207 + 118 = 325g
Mass of Pb(CH3COO)2 = 5g
Number of mole = Mass /Molar Mass
Number of mole of Pb(CH3COO)2 = 5/325 = 0.0154mole
From the equation,
1mole of Pb(CH3COO)2 produced 1mole of PbS.
Therefore, 0.0154mole of Pb(CH3COO)2 will also produce 0.0154mole of PbS
2. The name of the reaction is double displacement reaction since the ions in the two reactants interchange to form two different products
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
