Answer:
MgO + 2HBr → MgBr2 + H2O
Explanation:
Salt and water
Sugar and salt
Salt and pepper
Both trials of 1.2 g and 1.6 g will have the same mass percent of water because the ratio of the salt to the water of hydration is always constant for any hydrated salt.
<h3>Water of hydration</h3>
For every hydrated salt, the ratio of the salt to the water of hydration remains constant irrespective of the amount of salt taken for experimental analysis.
For example, assuming the mass percent of water in 10g of a hydrated salt is 40%, if 100g of the same salt is taken, the mass percent will remain 40%.
More on water of hydration can be found here: brainly.com/question/11202174
Answer:
39.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction
Ba(OH)₂(aq) + 2 CH₃COOH(aq) ⟶ Ba(CH₃COO)₂(aq) + 2 H₂O(l)
Step 2: Calculate the moles corresponding to 2.78 g of CH₃COOH
The molar mass of CH₃COOH is 60.05 g/mol.
2.78 g × 1 mol/60.05 g = 0.0463 mol
Step 3: Calculate the moles of Ba(OH)₂ needed to react with 0.0463 moles of CH₃COOH
The molar ratio of Ba(OH)₂ to CH₃COOH is 1:2. The moles of Ba(OH)₂ needed are 1/2 × 0.0463 mol = 0.0232 mol.
Step 4: Calculate the volume of 0.586 M solution that contains 0.0232 moles of Ba(OH)₂
0.0232 mol × 1 L/0.586 mol = 0.0396 L = 39.6 mL
