<span>The correct answer is b. Boiling point, why? because the liquid sample of napthalene is heated and remained at the temperature of 218 degrees celsius, the outcome was that the napthalene was completely vaporized, therefore we are given the scenario that at the temperature of 218 degrees celsius is considered to be the boiling pont of napthalene.</span>
thx but it's actually 143
Answer:
7.3 g (NH₄)₃PO₄
Explanation:
The balanced equation for the reaction is:
H₃PO₄ + 3 NH₃ ----> (NH₄)₃PO₄
To find the mass of ammonium phosphate ((NH₄)₃PO₄) produced, you need to (1) convert grams NH₃ to moles NH₃ (via the molar mass from the periodic table), then (2) convert moles NH₃ to moles (NH₄)₃PO₄ (via mole-to-mole ratio from balanced equation), and then (3) convert moles (NH₄)₃PO₄ to grams (NH₄)₃PO₄ (via molar mass from periodic table). Make sure to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs because the given value (2.5 grams) has 2 sig figs.
Molar Mass (NH₃): 14.01 g/mol + 3(1.008 g/mol)
Molar Mass (NH₃): 17.034 g/mol
Molar Mass ((NH₄)₃PO₄):
3(14.01 g/mol) + 12(1.008 g/mol) + 30.97 g/mol + 4(16.00 g/mol)
Molar Mass ((NH₄)₃PO₄): 149.096 g/mol
2.5 g NH₃ 1 mole NH₃ 1 mole (NH₄)₃PO₄ 149.096 g
--------------- x -------------------- x --------------------------- x --------------------------
17.034 g 3 moles NH₃ 1 mole (NH₄)₃PO₄
= 7.3 g (NH₄)₃PO₄
Answer:
Explanation:
Hello,
In this case, for the computation of the energy loss when the cooling process is carried out, we use the shown below equation:
Whereas we need the mass, specific heat and change in temperature of iron within the process. Thus, the only value we need is the specific heat that is 0.444 J/(g°C), therefore, we compute the heat loss:
Negative sign points out the loss due to the cooling.
Regards.
