The Answer to your question would be A
Calcium is used to isolate Rb from molten RbX because calcium has a smaller atomic radius than rubidium.
A chemical element's atomic radius, which is typically the average or typical distance between the nucleus's core and the outermost isolated electron, serves as a gauge for the size of an atom. There are numerous non-equivalent definitions of atomic radius since the border is not a clearly defined physical entity. Van der Waals radius, ionic radius, metallic radius, and covalent radius are the four most frequently used definitions of atomic radius. Atomic radii are typically measured in a chemically bound condition since it is challenging to isolated individual atoms in order to measure their radii individually.
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Balance each one by adding electrons to make the charges on both sides the same:
Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag
Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:
2 Ag+ + 2 e- --> 2 Ag
Now, just add the two half reactions together, cancelling anything that's the same on both sides:
2 Ag+ + Sn --> Sn2+ + 2 Ag
And you're done.
Answer: It's frequency also increases.
Explanation:
The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. ... The amplitude of a wave is related to the amount of energy it carries. A high amplitude wave carries a large amount of energy; a low amplitude wave carries a small amount of energy.
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Answer:
The frequency of photon is 0.75×10¹⁵ s⁻¹.
Explanation:
Given data:
Energy of photon = 5×10⁻¹⁹ J
Frequency of photon = ?
Solution:
Formula;
E = hf
h = planck's constant = 6.63×10⁻³⁴ Js
5×10⁻¹⁹ J = 6.63×10⁻³⁴ Js ×f
f = 5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js
f = 0.75×10¹⁵ s⁻¹
The frequency of photon is 0.75×10¹⁵ s⁻¹.
