Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
The number of bonds for a neutral atom is equal to the number of electrons in the full valence shell (2 or 8 electrons) minus the number of valence electrons. This method works because each covalent bond that an atom forms adds another electron to an atoms valence shell without changing its charge.
OILRIG:
Oxidation is loss (of electrons)
Reduction is gain (of electrons)
so...
The first one is an oxidation half-equation as the Sn loses electrons;
The second one is a reduction half-equation as the Cl₂ gains electrons
Answer:
2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.
Explanation:
The reaction:
<em>2H₂(g) + O₂(g) → 2H₂O(g), in presence of Pt as a catalyst.</em>
2.0 moles of hydrogen gas react with 1.0 mole of oxygen gas to produce 2.0 moles of water vapor in presence of Pt as a catalyst.
