Well in this
case, silver
nitrate is reduced:
Ag<span>+ </span><span>+ </span>e<span>− </span>→ Ag(s) ↓
Meanwhile, the aluminum
is oxidized forming a positive ion:
Al(s<span>) → </span>Al<span>3+ </span><span>+ 3</span>e−
To get the
overall reaction, we add the half
equations so that the electrons are eliminated:
Al(s<span>) + 3</span>Ag<span>+ </span><span>→ </span>Al<span>3+ </span><span>+ 3</span>Ag(s)
And similarly:
Al(s<span>) + 3</span>AgNO3(aq<span>) → </span>Al(NO3)3(aq<span>) + 3</span>Ag(s<span>)</span>
Answer:
This snip might help...it depends :)
Explanation:
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
<span>Answer is: the symbol is Cl.
[Ne ] 3s</span>² 3p⁶ is electric configuration of noble gas argon, neon (Ne) has10 electrons plus 6 electrons in 3s and 3p orbitals. Neutral atom of m<span>onatomic ion that has a charge of 1– has one electron less than argon, so that atom (chlorine) has 17 electrons. Charge of 1- means one electron more for ion: 17 + 1 = 18.
</span>
Answer:
compound
Explanation:
because now there are one together
