With a pH of five, it would mean that this substance is a weak acid one would also it expect it would be sour.
Answer:
What happens when it is squeezed is that its volume increases, the pressure of the material increases.
Explanation:
This is due to the fact that the elastic modulus of the sponge is high and withstands broad forces without deforming its structure, since the force is made within the proportional limit of its particles without modifying or permanently deforming them, that is why when stopping doing pressure or force on it its shape returns to being the original, this also happens due to the phenomenon of resilience
To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.
The balanced equation of the reaction will be:

Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +

So, the answer will be option A
(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
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>