Answer:
B) 3.0 g/mL
Explanation:
density formula: mass/volume
15/5=3
2 atoms of Al
3 of S
4*3=12 of O
so total= 17
When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.
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Answer:
a) IUPAC Names:
1) (<em>trans</em>)-but-2-ene
2) (<em>cis</em>)-but-2-ene
3) but-1-ene
b) Balance Equation:
C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄
As H₃PO₄ is catalyst and remains unchanged so we can also write as,
C₄H₁₀O → C₄H₈ + H₂O
c) Rule:
When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.
d) C is not Geometrical Isomer:
For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.
Molar mass of sodium chloride is 22.99 g/mol (Na+) + 35.45 g/mol (Cl-) = x g/mol. Then take the mass of sodium chloride sample 5 grams and divide by molar mass to get the number of moles. Take this number and multiply it by Avogadro’s number (6.22*10^23 molecules/mol). You now have the amount of molecules in 5 grams of NaCl.