<span>Which part of the electromagnetic spectrum has a higher frequency than ultraviolet light? </span>B. Gamma Rays.
Answer:
E = 15×10⁻²⁹ J
Explanation:
Given data:
Frequency of photon = 2.2× 10⁷ Hz
Energy of photon = ?
Solution:
Formula:
E = h.f
h = 6.63×10⁻³⁴ Js
by putting values,
E = 6.63×10⁻³⁴ Js × 2.2× 10⁷ s⁻¹
E = 14.586 ×10⁻²⁹ J
E = 15×10⁻²⁹ J
The energy of photon is 15×10⁻²⁹ J.
Positive ion with a radius smaller than the radius of the atom
Answer:
Ka2 values should be used in obtaining pKa2 in the Henderson–Hasselbalch equation
Explanation:
During first dissociation from H3PO4 to H2PO4-, ka1 will be used, and from H2PO4-to HPO4-, ka2 values will be used, hence a buffer solution containing H2PO4-and HPO4-will use ka2 in obtaining pKa2 in the Henderson–Hasselbalch equation.
The correct awser is the D/the 4th one
