They convert chemical potential energy (stored in muscle cells) to kinetic energy (running)
Gamma rays contain much more energy (most penetrating) than radio waves because they have a greater frequencies.
Radio waves are the electromagnetic waves with the longest wavelengths (1dm to 100 km), lowest frequencies (3kHz to 3GHz) and lowest energy (124 peV to <span>12,4 μeV).
</span>Gamma rays are the electromagnetic waves with the shortest wavelengths (1 pm), highest frequencies (300 EHz) and highest energy (1,24 Me<span>V</span>).
A chemical element that has an atomic number less than 58 and an atomic mass greater than 135.6m is barium (atomic no. 56 and atomic mass137.13 ) and lanthanum (atomic no. 57 and atomic mass 135.6).
<h3>Give a brief introduction about Barium and Lanthanum.</h3>
Barium is an element with the symbol Ba and atomic number 56. It is an alkaline earth metal that is soft and silvery, and it is the fifth element in group 2. Barium is never found in nature as a free element due to its extreme chemical reactivity. Oil well drilling fluid uses barium sulfate as an insoluble ingredient. It is employed as an X-ray radiocontrast agent in a purer form to image the human gastrointestinal tract. Barium compounds that dissolve in water have been employed as rodenticides despite being hazardous.
Chemical element lanthanum has the atomic number 57 and the symbol La. It is a silvery-white, ductile, soft metal that slowly tarnishes when exposed to air. It serves as the eponym for the group of 15 related elements in the periodic table between lanthanum and lutetium, of which lanthanum is the first and prototype. The rare earth elements traditionally include lanthanum.
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Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
