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Leto [7]
2 years ago
9

Determine whether the reaction will be spontaneous at high temperatures only, at low temperatures only, at all temperatures, or

no temperatures. (HINT: Use your chemical sense and your real-world knowledge to predict the signs of delta Hrxn & delta Srxn)
4Fe(s) + 3O2(g) ----> 2Fe2O3(s) [rust]


Circle one:High T, Low T, All T, No T
Chemistry
1 answer:
LiRa [457]2 years ago
5 0

Answer:

The rusting of iron is spontaneous at low temperatures.

Explanation:

The given chemical reaction is:

4Fe(s) + 3O2(g) ----> 2Fe2O3(s) [rust]

The rusting of iron is a chemical reaction in which iron reacts with oxygen in presence of moisture and forms iron oxide.

This reaction takes place in a faster rate when there is low temperatures in the atmosphere.

When temperature is low, the moisture in the atmosphere is more and hence, rate of rusting is more.

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Which has a greatest mass, one atom of carbon, one atom of hydrogen, or one atom of litium
zysi [14]

Answer:

One atom of carbon

Explanation:

5 0
2 years ago
Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give th
natta225 [31]

Answer:

1. Orbital diagram

2p⁴   ║ ↑↓ ║  "↑"  ║   ↑

2s²    ║ ↑↓ ║

1s²     ║ ↑↓ ║

2. Quantum numbers

  • <em>n </em>= 2,
  • <em>l</em> = 1,
  • m_{l} = 0,
  • m_{s} = +1/2

Explanation:

The fill in rule is:

  • Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2
  • Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
  • Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴   ║ ↑↓ ║  "↑"  ║   ↑

2s²    ║ ↑↓ ║

1s²     ║ ↑↓ ║

The quantum number of an electron consists of four number:

  • <em>n </em>(shell number, - 1, 2, 3...)
  • <em>l</em> (subshell number or  orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
  • m_{l} (orbital energy, or "which box the electron is in"). For example, orbital <em>p </em>(<em>l</em> = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital <em>d</em> (<em>l </em>= 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
  • m_{s} (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number

  • <em>n </em>= 2, shell number 2,
  • <em>l</em> = 1, subshell or orbital <em>p,</em>
  • m_{l} = 0, 2nd "box" in the range -1, 0, 1
  • m_{s} = +1/2, single electron always has +1/2
3 0
3 years ago
A hydrogen-like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by the
yan [13]

Answer:

The ionization energy (in kJ/mol) of the helium ion is 21,004.73 kJ/mol .

Explanation:

E_n = -(2.18 10-18 J)\times \frac{Z^2}{n^2}

Z = atomic mass

n = principal quantum number

Energy of the electron in n=1,

E_1= -(2.18 10^{-18} J)\times \frac{4^2}{1^2}=-3.488\times 10^{-17} J

Energy of the electron in n = ∞

E_{\infty}= -(2.18 10^{-18} J)\times \frac{2^2}{\infty ^2}=0 J

Ionization energy of the He^+ ion:

I.E=E_{infty}-E_1=0-(-3.488\times 10^{-17} J)=3.488\times 10^{-17} J

I.E=3.488\times 10^{-20} kJ

To convert in into kj/mol multiply it with N_A=6.022\times 10^{23} mol^{-1}

I.E=3.488\times 10^{-20} kJ\times 6.022\times 10^{23} mol^{-1}=21,004.73kJ/mol

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