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vova2212 [387]
3 years ago
9

Which is a defining trait of a mineral?

Chemistry
1 answer:
Orlov [11]3 years ago
8 0
Traits  which are used to identify and describe mineral species.
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"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
3 years ago
How many dm³ of hydrogen,measured at s.t.p.,would be needed to reduce 47.7g of copper(II) oxide to copper?
GaryK [48]

Answer:

Option D. 13.44

Explanation:

We'll begin by calculating the number of mole in 47.7g of copper(II) oxide, CuO.

This can be obtained as follow:

Mass of CuO = 47.7 g

Molar mass of CuO = 63.5 + 16 = 79.5 g/mol

Mole of CuO =.?

Mole = mass /Molar mass

Mole of CuO = 47.7/79.5

Mole of CuO = 0.6 mole

Next, we shall write the balanced equation for the reaction. This is given below:

CuO + H2 —> Cu + H2O

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2 to produce 1 mole of Cu and 1 mole of H2O.

Next, we shall determine the number of mole of H2 needed to react completely with 0.6 mole of CuO.

This can be obtained as follow:

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2.

Therefore, 0.6 mole of CuO will also react with 0.6 mole of H2.

Finally, we shall determine the volume occupied by 0.6 mole of H2 at STP.

This can be obtained as follow:

1 mole of H2 occupied 22.4 dm³ at STP.

Therefore, 0.6 mole of H2 will occupy = 0.6 × 22.4 = 13.44 dm³.

Therefore, 13.44 dm³ of H2 is needed for the reaction.

4 0
2 years ago
What process is used to release energy in nuclear power plants?
lukranit [14]
Nuclear fission is used enable to release energy in power plants. The constant collision of particles within the reactor, create most of the plants energy.
6 0
3 years ago
Read 2 more answers
Complete the following single replacement reaction. If they don’t react, just write “NR”
stiv31 [10]

We have to complete all the given reactions.

1. Fe(s) + CuCl₂ → Cu + FeCl₂

2. Cu(s) + FeCl₂(aq)  → NR (no reaction takes place)

3. K(s) + NiBr2(aq) → NR (no reaction takes place)

4. Ni(s) + KBr(aq) → K + NiBr₂

5. Zn(s) + Ca(NO₃)₂(aq) → NR (no reaction)

6. Ca(s) + Zn(NO₃)₂(aq) → Zn(s) + Ca(NO₃)₂(aq)

4 0
3 years ago
Use the following information to answer the following question.
Irina-Kira [14]

Answer:

For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%  

Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver  

12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:  

38.1-g/43.2-g x 100% = 88.2%

Explanation:

5 0
2 years ago
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