Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
B and e
first we need to balance the NH3 hence first we do E and multiplying the coefficient by 2. that will leave us with N2+H2–>2NH3.
N2 and H2 is balanced and now all that is left to do is the balance H2 by 3 as there is 6H on RHS hence we need 6H on LHS
Enthalpy change during the dissolution process = m c ΔT,
here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
= -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)
Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>
