You will need the Gas Law:
pV = nRT
Since T and p are constant, R is constant too, then moles increases->volume will increase with the same ratio too!
Answer:
1.47 atm
Explanation:
Step 1: Calculate the moles corresponding to 41.6 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
41.6 g × 1 mol/32.00 g = 1.30 mol
Step 2: Convert 30.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 30.0 + 273.15 = 303.2 K
Step 3: Calculate the pressure exerted by the oxygen
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 1.30 mol × (0.0821 atm.L/mol.L) × 303.2 K / 22.0 L = 1.47 atm
Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
Answer:
D. An acid accepts an electron pair from a base.
Explanation:
acorroding to lewis
- an acid is a compound which accepts a pair of electrons from base .
- a base is a compound which donates a pair of electron to acid .
