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goldfiish [28.3K]
3 years ago
6

Name the noble gas atom that has the same electron configuration as each ion in the compound rubidium oxide

Chemistry
1 answer:
Zepler [3.9K]3 years ago
3 0
In rubidium oxide - Rb₂O , the ions are Rb⁺ and O²⁻
Rb is a group one element with one valence electron. To become stable it loses its outer electron to gain a complete outer shell.
Electronic configuration of Rb is - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
Once it loses its valence electron the configuration is;
 - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 
The noble gas with this configuration is Krypton - Kr

Oxygen electron configuration is 1s² 2s² 2p⁴
Once it gains 2 electrons the configuration is  - 1s² 2s² 3p⁶
The noble gas with this configuration is Neon - Ne
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3 years ago
How long would it take for 1.50 mol of water at 100.0 âc to be converted completely into steam if heat were added at a constant
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To convert boiling water to steam, that would involve heat of vaporization. The heat of vaporization for water at atmospheric conditions is: ΔHvap = <span>2260 J/g.

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Q = m</span>ΔHvap = (1.50 mol water)(18 g/mol)(<span>2260 J/g) = 61,020 J

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6 0
2 years ago
Given that Kp [NOTE: Kp!!!!] = 1.39 at 400 ºC for the reaction, P4(g) &lt;=&gt; 2 P2(g), which answer best describes the reactio
Igoryamba

Answer:

The reaction will proceed to the left to attain equilibrium.

Explanation:

The question is missing but I guess it must be about <em>how the reaction will proceed to attain equilibrium.</em>

First, we have to calculate the partial pressures using the ideal gas equation.

pP_{4}=\frac{2.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L} =5.52atm

pP_{2}=\frac{1.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L}=3.31atm

Now, we have to calculate the reaction quotient (Qp).

Qp=\frac{pP_{2}^{2}}{pP_{4}} =\frac{3.31^{2} }{5.52} =1.98

Since Qp > Kp, the reaction will proceed to the left to attain equilibrium.

3 0
2 years ago
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