Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Answer:
a. HCl.
b. 0.057 g.
c. 1.69 g.
d. 77 %.
Explanation:
Hello!
In this case, since the reaction between magnesium and hydrochloric acid is:

Whereas there is 1:2 mole ratio between them.
a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.
b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

Thus, the mass of excess magnesium turns out:

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

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Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³