A gamma ray has a mass and atomic number of 0 meaning it doesn’t effect the nucleus.
E.g
60 60 0
Co —> Co + gamma ray
27 27 0
Answer:
The balanced half-reaction in which sulfate ion is reduced to sulfite ion is a two electron process
Explanation:
In the sulfate anion, S, acts with +6 in the oxidation number
SO₄⁻²
This is the anion from the sulfuric acid, when it lost the 2 protons
H₂SO₄ → 2H⁺ + SO₄⁻²
In the sulfite anion, S, acts with +4 in the oxidation number.
SO₃⁻²
This is the anion, generated when the sulfurose acid lost the 2 protons.
H₂SO₃ → SO₃⁻² + 2H⁺
In acidic medium
2H⁺ + SO₄⁻² + 2e⁻ → SO₃⁻² + H₂O
In basic medium
H₂O + SO₄⁻² + 2e⁻ → SO₃⁻² + 2OH⁻
Active metals are those metals in the group 1 of the periodic table.
Electronegativity is the trend to atract electrons.
Active metals have few valence electron, because their last shell is of the kind ns^1 or ns^2
Then, these atoms do not trend to attract electrons. The most electronegative atomos are those who have 7 valence elecfrons; this is their last shell is of the kind ns^7, because when they attract one electron to its valence shell they will complete 8 electrons which is the most stable configuration.
The silver ion concentration in saturated solution of silver (i) phosphate is calculated as follows.
write the equation for dissociation of silver (i) phosphate
that is Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)
let the concentration of the ion be represented by x
ksp is therefore= (3x^3 )(x) = 1.8 x10 ^-18
27 x^3 (x) = 1.8 x10^-18
27x^4 = 1.8 x10^-18 divide both side 27
X^4 = 6.67 x10 ^-20
find the fourth root x = 1.6 x10 ^-5m
the concentration of silver ion is therefore = 3 x (1.6 x10^-5) = 4.8 x10^-5m
