.the sum of the pressures that each gas would exert if they occupied twice the volume
Solution :
Comparing the solubility of silver chromate for the solutions :
----- Less soluble than in pure water.
----- Less soluble than in pure water.
----- Similar solubility as in the pure water
----- Similar solubility as in the pure water
The silver chromate dissociates to form :

When 0.1 M of
is added, the equilibrium shifts towards the reverse direction due to the common ion effect of
, so the solubility of
decreases.
Both
and
are neutral mediums, so they do not affect the solubility.
(4) chemical energy to electrical energy is the correct answer.
Hope this helps~
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of
) =-1275.0
enthalpy of combustion of oxygen(Δ
of
) = zero
enthalpy of combustion of carbon dioxide(Δ
of
) = -393.5
enthalpy of combustion of water(Δ
of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ
= ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ
= [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ
= [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ
= 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ
= -2361 - 1714 - 0 + 1275
Δ
=-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>