If iodine is added to a starch solution, they react with each other and the iodine darkens to an almost pitch black.
however, if iodine is added to a solution containing no starch, it will show up only as an extremely pale brown. almost colorless and hardly visible.
when following the changes in some inorganic oxidation reduction reactions, iodine may be used as an indicator to follow the changes of iodide ion and iodine element. soluble starch solution is added. only iodine element in the presence of iodide ion will give the characteristic blue black color. neither iodine element alone nor iodide ions alone will give the color result.
hope this answer really helps your question :)
The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
<h3>X-Rays contradict to?</h3>
<h3>C. gamma </h3>
a type of penetrating electromagnetic radiation produced by the radioactive disintegration of atomic nuclei
The magnetic field at the center of the arc is 4 × 10^(-4) T.
To find the answer, we need to know about the magnetic field due to a circular arc.
<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
- According to Biot savert's law, magnetic field at the center of a circular arc is
- B=(μ₀ I/4π)× (arc/radius²)
- As arc is given as angle × radius, so
B=( μ₀I/4π)×(angle/radius)
<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>
B=(μ₀ I/4π)× (0.9/0.006)
= (10^(-7)× 26.9)× (0.9/0.006)
= 4 × 10^(-4) T
Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.
Learn more about the magnetic field of a circular arc here:
brainly.com/question/15259752
#SPJ4