M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
#1. An element or ion that has lost two electrons must have a net charge of 2+, because it has two more protons than electrons, therefore the answer is Mg2+
#2. aluminum ions have an oxidation state of 3+ and fluoride has an oxidation state of 1-, therefore I’m order for the charges to cancel you need 3 fluoride ions.
Therefore, the answer is AlF3
I believe it's about the dislodging response as Ag structures solvent salt with nitrate. For dislodging response, more receptive metal will uproot less receptive metal from the arrangement. In your question, the reactivity of the metal are positioned as takes after: Mg > Cu > Ag. Take note of that more responsive metals, that are Mg and Cu, are in the arrangement.
16) Na (s) + H2O(L) ---> H2 (g) + NaOH (aq)
17) O2 (g) + NH3 (g) --->H2O (L) + HNO3 (aq)
18) K (s) + Cl2 (g) ---> KCl (s)
19) Al (s) + HCl (aq) ---> H2 (g) + AlCl (aq)
20) Na3PO4 (aq) + CaCl2 (aq) ---> NaCl (s) + Ca3(PO4)2 (s)
