"I" symbol means the current goes through the system (imagine the 'I' being a line, like a circuit connecting [power to the device]) "O" symbol means the current does not go through the system. ( the circle is an open circuit, having no power flowing through it
it is nutrients that's it
When sulfur reacts with zinc metal is forms a compound known as zinc sulfide.
<h3>
Name of the elements</h3>
The names of the elements that make up the compound will be determined from the atomic mass of the elements.
<h3>Which element has atomic number 30</h3>
The element that atomic number of 30 is zinc and it has oxidation number of +2.
<h3>Which element has atomic number 16</h3>
The element that atomic number of 16 is sulfur and it has oxidation number of -2.
<h3>Chemical reaction between sulfur and zinc metal</h3>
Zn²⁺ + S²⁻ --------> ZnS
Thus, when sulfur reacts with zinc metal is forms a compound known as zinc sulfide.
Learn more about zinc sulfide here: brainly.com/question/20806552
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Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
