O2 is the limiting reactant
Answer:
29.2 L
Explanation:
Given data

We can find the volume of the gas (V) of the using the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 1 mol × (0.0821 atm.L/mol.K) × 293 K / 0.824 atm
V = 29.2 L
I think it's D but not sure
Answer:
The molar mass of Mg(NO₃)₂, 148.3 g/mol.
Explanation:
Step 1: Given data
- Mass of Mg(NO₃)₂ (solute): 42.0 g
- Volume of solution: 259 mL = 0.259 L
Step 2: Calculate the moles of solute
To calculate the moles of solute, we need to know the molar mass of Mg(NO₃)₂, 148.3 g/mol.
42.0 g × 1 mol/148.3 g = 0.283 mol
Step 3: Calculate the molarity of the solution
M = moles of solute / liters of solution
M = 0.283 mol / 0.259 L
M = 1.09 M