Answer:

Explanation:
Hello,
In this case, the reaction is given as:

Thus, starting by the yielded grams of silver iodide, we obtain:

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.
Best regards.
Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.
In this problem, we have density and we have mass so we can plug into the equation and solve for V.
38.6=270.2/V
<em>*Multiply both sides by V*</em>
38.6V=270.2
<em>*Divide both sides by 38.6*</em>
V=7
The volume of the gold nugget is 7cm3.
Hope this helps!!
Calcium carbon oxygen, calcium chlorine, carbon hydrogen oxygen, hydrogen chlorine, calcium oxygen hydrogen I think ♀️ Hopefully it’s right or I helped a bit
Answer:
Solubilities of
increases with increasing acidity
Explanation:

Concentration of
decreases with increasing acidity. Therefore solubility of
increases to keep constant solubility product.

Concentration of
decreases with increasing acidity due to formation of
. therefore solubility of ZnS increases to keep constant solubility product

Increasing acidity have no effect on this above equilibrium. Therefore solubility of
remains constant.

Concentration of
decreases with increasing acidity due to formation of
. Therefore solubility of
increases to keep constant solubility product.

Increasing acidity have no effect on this above equilibrium. Therefore solubility of NaI remains constant.

Concentration of
decreases with increasing acidity due to formation of
. Therefore solubility of
increases to keep constant solubility product.