The element iodine (I) is important for the fast and hastened metamorphosis of frog-tadpoles. As amphibians, the tadpoles can live in water and land but when they are born they are iodine-deficient. Tadpoles that do not receive ample amount of iodine become tadpoles until the end of their days.
Answer :
The steps involved in the electron dot structure of
are :
First we have to determine the total number of valence electron in
.
Number of valence electrons in N = 5
The charge on N is (-3). So, we add 3 electrons.
Total number of valence electrons = 5 + 3 = 8 electrons
The image is shown below.
<em>answer:</em><em> </em><em>option </em><em>d </em><em>(</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>H </em><em>+</em><em>2</em><em>×</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>mass </em><em>of </em><em>O</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
<span>(a) what is the average volume (in cubic meters) required for each iron atom
For this case, the density of Iron would be </span>7.87g/cm³
<span>
V = 9.27 x 10^-26 kg / </span>7.87g/cm<span>³ ( 1 kg / 1000 g)
</span>V = 1.18 x 10-23 cm³<span>
(b) what is the distance (in meters) between the centers of adjacent atoms?
We assume the atoms as cube, so we use the volume of the cube to calculate the distance of the atoms.
V = </span>1.18 x 10-23 cm<span>³ = s</span>³
s = 2.28 x 10^-8 cm