9. How does the electron configuration of an ion of

Chemistry

1 answer:

madreJ [45]1 year ago

4 0

Explanation:

Group 1A consist of metals and group. 2A consist of non metals . group1A elements can become stable by donating 1 electron in the same way the elements of group 7A also becomes stable by gaining one electron . So due to being so close to being a inert gas element it is compared to the nearest noble gas.

You might be interested in

Determine the energy in joules of a photon whose frequency is 3.55 x10^17 hz( with units)

kogti [31]

i think it is this

Explanation:

3 0

2 years ago

3. 25. The temperatures favorable for the growth of most bacteria are

kvasek [131]

about the same as for humans

Explanation:

This is why most bacteria, during research in labs, are incubated at 37 degrees centigrade, about the same as human body temperature. In addition, harmful and beneficial bacteria thrive in the human body due to the favorable temperatures for growth and reproduction. To try and fight an infection, the body also tries to raise the body temperatures above the optimal for the bacteria growth (the reason one has a fever in case of infection).

6 0

3 years ago

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$