Answer:
the Molar heat of Combustion of diphenylacetylene 
= 
Explanation:
Given that:
mass of diphenylacetylene = 0.5297 g
Molar Mass of diphenylacetylene = 178.21 g/mol
Then number of moles of diphenylacetylene = 
= 
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene = Heat absorbed by 
+ Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene = msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene = 20598.87 J
Heat liberated by 1 mole of diphenylacetylene will be = 
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene =
Use the question marck Moles of CO2
The the giving = 0.624 mol O2
Find the CF faction = 1 mole= 32.00 of O2
O= 2x16.00= 32.00amu ( writte this in the cf fraction)
SET UP THE CHART
Always start with the giving
0.624 mol O2 / 1mol of CO2
___________ / _____________ = Cancel the queal ( O2)
/ 32.00c O2
/
/
Multiply the top and divide by the bottom
0.624 mol CO x 1mol CO2 = 0.624 divide by 32.00 O2 =0.0195
You should look at the giving number ( how many num u gor ever there)
Ur answer should have the same # as ur givin so
= 0.0195
= .0195 mol of CO2
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
32.8 C
Explanation:
- Use combined gas law formula and rearrange.
- Hope that helped! Please let me know if you need further explanation.
Cycloalkanes are those saturated organic compounds which exist in the form of Rings. Their Hydrogen Deficiency Index in one. The General formula for cycloalkanes is,
CnH2n
When number of Carbons = 8
Then
C₈H₂₍₈₎
C₈H₁₆
Result:
Cycloalkane containing 8 carbon atoms has
16 hydrogen atoms.
