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2 years ago

What is the electron configuration for iodine

Chemistry

1 answer:

iogann1982 [59]2 years ago

7 0

Hello!

To find electron configuration for Idoine we need to understand the following steps:

Finding the Atom's Atomic Number (tells us the specific number of electrons)
Determining the Charge of the Atom
Understanding the orbitals (Set S [Contains 2], P [Contains 3, Holds 6], D [Contains 5, Holds 10], F [Contains 7, Holds 14], and there are some theoritical ones.) [Overall the sets go SPDFGHIK
Understanding notations in configuartion. The notations display the number of electrons in the atom and set.

In this case, for Iodine. If we follow these rules we can see that the electron configuration is [Kr] 4d^10 5s^2 5p^5. We use Krytpon in front because that is the last full and stable noble gas before this particular element. Atoms are just trying to be stable so the goal is to achieve that full shell.

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C(s) + S8(s) → CS2(l)

Vanyuwa [196]

The balanced chemical reaction is written as:

4C(s) + S8(s) → 4CS2(l)

We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.

7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8

The limiting reactant would be S8. We use this amount to calculate.

0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2

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3 years ago

Read 2 more answers

In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz,

Vika [28.1K]

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

$\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}$

$\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}$

$Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz$

Therefore, the signal is at 11759 Hz from the TMS.

6 0

3 years ago