Answer:
When an object is heated up, its particles gain more kinetic energy and hence its internal energy increases. A change in state will also result in a change in internal energy.
When an object cools down, heat is withdrawn from it. Hence, the entropy of the object decreases. But the decrease in entropy leads to the transfer of energy to the surrounding.
I hope this helps you! Good luck with school! :D
The statement "<span>The liquid-vapor equilibrium of a substance can be disturbed by cooling the substance" is false. The substance must be heated in order to get the vapor phase of that substance.</span>
Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"
Answer:
Fe₂O₃ is the limiting reactant.
7.57 g of MgO are formed.
Explanation:
- 3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO
First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 15.6 g Mg ÷ 24.305 g/mol = 0.642 mol Mg
- 10.0 g Fe₂O₃ ÷ 159.69 g/mol = 0.0626 mol Fe₂O₃
0.0626 moles of Fe₂O₃ would react completely with (3 * 0.0626 ) 0.188 moles of Mg. As there are more Mg moles than required, Mg is the reactant in excess; thus, <em>Fe₂O₃ is the limiting reactant</em>.
We now <u>calculate how many MgO moles are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.0626 mol Fe₂O₃ *
= 0.188 mol MgO
Finally we <u>convert moles of MgO into grams</u>:
- 0.188 mol MgO * 40.3 g/mol = 7.57 g
