Answer:
Na.
Explanation:
- The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).
- An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced.
- A reducing agent (also called a reductant or reducer) is an element (such as calcium) or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction.
<em>2Na + S → Na₂S.</em>
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Na is oxidized to Na⁺ in (Na₂S) (loses 1 electron). "reducing agent".
S is reduced to S²⁻ in (Na₂S) (gains 2 electrons). "oxidizing agent".
Sunlight, soil, water, leaves,
Addition of boiled, deionized water to the titrating flask to wash the wall of the erlenmeyer flask and the buret tip will have no effect on the Ksp value of ca(oh)2.
There will be no effect on the Ksp value as boiled deionised water is not able to alter the number of hydronium and hydroxide ions. As no change in the ions happen so there will be no change in Ksp value. The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, Ksp. It stands for the degree of solute dissolution in solution. A substance's Ksp value increases with how soluble it is.
To know more about, solubility product constant, click here,
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Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
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ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>