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viva [34]
3 years ago
15

Draw the Lewis dot structure for the compound gallium oxide with the chemical formula Ga2O3

Chemistry
1 answer:
Brut [27]3 years ago
3 0

Answer:

Refer to your periodic table. Lewis dot structures are based off the number of valence electrons an atom has.

Looking at the compounds, we can see that Gallium has three valence electrons in its outer shell and oxygen has six. Oxygen and Gallium are going to share electrons with one another, making a V shape in their diagram.

One Oxygen would make a double bond with a Gallium, leaving one valence electron to another oxygen. That oxygen takes that Final electron. It now has 7 in its outer shell. The remaining Gallium and Oxygen do the same double bond as the one before, leaving the 7 valence electron oxygen with one more electron.

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3 0
3 years ago
Read 2 more answers
4. Which of the following NOT TRUE about real gas
Semenov [28]

Answer:

The answer is D.

Explanation:

Intermolecular force are negligible

When the distance between molecules decrease,

the attraction or repulsion become greater

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2 years ago
Describes liquids that do not dissolve<br> in each other
gregori [183]

Answer:

Immiscible

Explanation:

8 0
3 years ago
The reaction AB(aq)→A(g)+B(g) is second order in AB and has a rate constant of 0.0164 M −1 ⋅ s −1 at 25.0 ∘ C . A reaction vesse
uysha [10]
The reaction is second order in AB, so: v=k[AB]^2. In the statement, we obtain that [AB]=0.104~M and, at 25 ºC, k=0.0164~M^{-1}\cdot s^{-1}. Then:

v=k[AB]^2\\\\&#10;v=0.0164\cdot0.104^2\\\\&#10;v=0.0164\cdot0.010816\\\\&#10;v\approx0.000177=1.77\times10^{-4}~mol/s

Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:

\bullet~\text{Pressure:}~p=707.3-23.8=683.5~mmHg\\\\&#10;\bullet~\text{Volume:}~V=142~mL=0.142~L\\\\&#10;\bullet~\text{Number of moles:}~n=n_A+n_B\\\\&#10;\bullet~\text{Ideal gas constant:}~R=62.3~L\cdot mmHg\cdot K^{-1}\cdot mol^{-1}\\\\&#10;\bullet~\text{Temperature:}~T=25^oC=25+273~K=298~K\\\\

pV=nRT\\\\&#10;683.5\cdot0.142=n\cdot62.3\cdot298\\\\&#10;n=\dfrac{683.5\cdot0.142}{62.3\cdot298}\\\\&#10;n\approx0,0052~mol

Since each AB molecule forms one of A and one of B, n_A=n_B. Hence: 2n_A\approx0,0052\Longrightarrow n_A=n_B\approx0.0026~mol.

We'll consider that in the beginning there was not A or B. So, \Delta n_A=\Delta n_B=0.0026-0=0.0026~mol. Furthermore, since the ratio of AB to A and to B is 1:1, |\Delta n_{AB}|=|\Delta n_A|=|\Delta n_B|.

Calculating the time by the expression of velocity:

v=\dfrac{|\Delta[AB]|}{\Delta t}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_{AB}|}{V}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_A||}{250~mL}\\\\&#10;1.77\cdot10^{-4}=\dfrac{1}{\Delta t}\cdot\dfrac{0.0026~mol}{0.25~L}\\\\&#10;\Delta t=\dfrac{0.0026}{0.25\cdot1.77\cdot10^{-4}}\\\\&#10;\boxed{\Delta t\approx58.76~s}
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