Answer:
24.4 amu or g/mole
Explanation:
24 x 0.790 = 19.0 amu
25 x 0.100 = 2.50 amu
26 x 0.110 = 2.86 amu
(Because of the 19.0, the sig figs go only to the 1/10 decimal place)
19.0 + 2.5 + 2.9 = 24.4 amu or g/mole
<h3>
Answer:</h3>
Al- [Ne] 3s²3p¹
As- [Ar] 4s²3d¹⁰ 4p³
Explanation:
- Electron configuration of an element shows the arrangement of electrons in the energy levels or orbitals in the atom.
- Noble-gas configuration involves use of noble gases to write the configuration of other elements.
- This is done by identifying the atomic number of the element and then identifying the noble gas that comes before that particular element on the periodic table.
- Aluminium: The atomic number of Al is 13. The noble gas before Aluminium is Neon which has 10 electrons. Therefore the remaining 3 electrons fills up the 3s and 3p sub orbitals.
- Thus, the noble-gas configuration of Al is [Ne] 3s²3p¹
2. Arsenic, Atomic number is 33
- Noble gas before Arsenic is Ar,. Argon has 17 electrons, then the remaining electrons fills up the 4s, 3d and 4p sub-orbitals.
- Thus, the noble-gas configuration of As is [Ar] 4s²3d¹⁰ 4p³
Answer:
Half-reactions:
Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻
Net ionic equation:
2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺
Explanation:
The Cr³⁺ is reduced to Cr²⁺:
<h3>
Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>
Zn is oxidized to Zn²⁺:
<h3>
Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>
Twice the reduction of Cr:
2Cr³⁺ + 2e⁻ → 2Cr²⁺
Now this reaction + Oxidation of Zn:
2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻
<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>
Answer: Its A or D
wish i had an actual answer sorry..
