Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given solution is as follows.
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Answer: 9.18 Litres
Standard Temperature and Pressure (STP). Think of this as the perfect environment where the Temp. is 0°C or 273 Kelvin and Pressure is always 1 atm. This is only true in STP.
This question uses the Ideal Gas Equation:
PV=nRT
P= 1 atm
V = ??
T = 273 K (always convert to Kelvin unless told otherwise)
n = 0.410 mol
R = 0.0821 L.atm/mol.K
What R constant to use depends on the units of the other values. (look at the attachments) The units cancel out and only Litres is left. You simply multiply the values.
