Answer:
We have to weigh 52.8 g of BaCl₂·2H₂O, add it to a 2.00 L flask and add water until reaching the final volume.
Explanation:
<em>Describe the preparation of 2.00 L of 0.108 M BaCl₂ from BaCl₂·2H₂O. (244.3 g/mol).</em>
Step 1: Calculate the moles of BaCl₂
We need to prepare 2.00 L of a solution that contains 0.108 moles of BaCl₂ per liter of solution.
2.00 L × 0.108 mol/L = 0.216 mol
Step 2: Calculate the moles of BaCl₂·2H₂O that contain 0.216 moles of BaCl₂
The molar ratio of BaCl₂·2H₂O to BaCl₂ is 1:1. The moles of BaCl₂·2H₂O required are 1/1 × 0.216 mol = 0.216 mol.
Step 3: Calculate the mass corresponding to 0.216 mol of BaCl₂·2H₂O
The molar mass of BaCl₂·2H₂O is 244.3 g/mol.
0.216 mol × 244.3 g/mol = 52.8 g
We have to weigh 52.8 g of BaCl₂·2H₂O, add it to a 2.00 L flask and add water until reaching the final volume.
Answer:
B
Explanation:
Crush solid reactant into smaller pieces.
The answer to this question is D
Answer:
c
Explanation:
all the atoms must be balanced.
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
