Formula for Barium Nitrate = Ba(NO3)2
Thus based on stoichiometry:
1 mole of Ba(NO3)2 contains 2 moles of NO3-
Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M
Use the relation:
V1M1 = V2M2
V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L
Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.
Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Explanation:
According to ideal gas equation:
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 0.684
R = gas constant = 
T =temperature =
(at STP)
Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
The molar volume of the gas : V = 447.993 L
<h3>Further explanation</h3>
Given
n = 1 mol
P = 0.05 atm
T = 273 K
Required
The molar volume
Solution
Ideal gas law :
PV=nRT
V = nRT/P
Input the value :
V = 1 x 0.08205 x 273 / 0.05
V = 447.993 L
It looses 2 electrons, when ions come together to form and ionic bond they always join in numbers that cancel out the positive and negative charges .
-hope this have helped you well .
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