First there is a need to calculate the molar mass of Ba(NO₃)₂:
137.3 + 2 (14.0) + 6 (16) = 261.3 grams/mole
The molar mass, denoted by M in chemistry refers to a physical characteristic illustrated as the mass of a given component divided by the amount of the component. The molar masses are always denoted in grams/mole.
After finding the molar mass, the number of moles can be identified as:
432 grams / 261.3 g/mol = 1.65 moles of Ba(NO₃)₂.
Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
Answer: [D]: "They are not part of the rock cycle." .
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Answer:
We have to take 37.5 mL of a 0.400 M solution
Explanation:
Step 1: Data given
Stock volume = 100 mL = 0.100L
Stock concentration 0.400 M
Volume of solution he wants to make = 100 mL = 0.100L
Concentration of solution he wants to make = 0.150 M
Step 2: Calculate the volume of 0.400 M CuSO4 needed
C1*V1 = C2*V2
⇒with C1 = the stock concentration = 0.400M
⇒with V1 = the volume of the stock = TO BE DETERMINED
⇒with C2 = the concentration of the solution he wants to make = 0.150 M
⇒with V2 = the volume of the solution made = 0.100 L
0.400 M * V1 = 0.150M * 0.100L
V1 = (0.150M*0.100L) / 0.400 M
V1 = 0.0375 L = 37.5 mL
We have to take 37.5 mL of a 0.400 M solution
