Rutherford's gold foil experiment proved that there was a small, dense, positively charged nucleus at the center, which contained most of the mass of the atom. Which contained electrons orbiting the nucleus.
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:
Molecular formula = C₄H₆As₆Cu₄O₁₆
Explanation:
Given data:
Empirical formula = C₂H₃As₃Cu₂O₈
Molar mass of compound = 1013 g/mol
Molecular formula = ?
Solution:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass of C₂H₃As₃Cu₂O₈ is 506.897 g/mol
by putting values.
n = 1013 / 506.897
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₂H₃As₃Cu₂O₈)
Molecular formula = C₄H₆As₆Cu₄O₁₆
P1: 741 mmHg
V1: 3.49 L P1 x V1 / P2= (741 mmHg) (3.49 L) / 760 mmHg = 3.40 L
P2: 760 mmHg
V2: ? L
