What part of an atom has most of its mass

"Write the electron Configuration expected for element 113 and the configurations for the two cations it is most likely to form"

arlik [135]

Answer:

Element:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Cations:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

Explanation:

The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.

So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

5 0

2 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

5 0

2 years ago

Why is the mass of an iron atom different from the mass of a copper atom?

Jobisdone [24]

They have different number of Neutrons and protons, so their masses are different

Hope this helps!

6 0

3 years ago

21,980 mg of phosphorus reacts in a gas jar containing oxygen. The resulting oxide weighed 50.000 g. The percent composition of

stealth61 [152]

Answer: False

Explanation:

8 0

3 years ago

An irregular chunk of iron has a density of 7.87 g/cm3. A 14.5 gram chunk is added to a graduated cylinder. If the final volume

bulgar [2K]

Answer:

37.96 mL

Explanation:

From the question given above, the following data were obtained:

Density (D) of iron = 7.87 g/cm³

Mass (m) of iron = 14.5 g

Volume of Cylinder + Iron = 39.8 mL

Volume of Cylinder =?

Next, we shall determine the volume of the chunk of iron. This can be obtained as follow:

Density (D) of iron = 7.87 g/cm³

Mass (m) of iron = 14.5 g

Volume (V) of iron =?

D = m/V

7.87 = 14.5 /V

Cross multiply

7.87 × V = 14.5

Divide both side by 7.87

V = 14.5 / 7.87

V = 1.84 cm³

Recall:

1 cm³ = 1 mL

Therefore,

1.84 cm³ = 1.84 mL

Therefore, the volume of the chunk of iron is 1.84 mL.

Finally, we shall determine the starting volume of the cylinder as follow:

Volume of Cylinder + Iron = 39.8 mL

Volume of Iron = 1.84 mL.

Volume of Cylinder =?

Volume of Cylinder = (Volume of Cylinder + Iron) – Volume of Iron

Volume of Cylinder = 39.8 – 1.84

Volume of Cylinder = 37.96 mL

Therefore the starting volume of cylinder is 37.96 mL

3 0

2 years ago