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2 years ago

Please help with the question

Chemistry

1 answer:

Ludmilka [50]2 years ago

3 0

Answer: NO2.

You can predict that two atoms with similar electronegativity will form covalent compounds while atoms with pretty different electronegativity will form ionic compounds.

Given that N and O are neighbors on the periodic table, you can predict they have similar electronegativities.

Also you can predict that from the fact that N and O are nonmetals, which also means that they will form covalent compounds.

The other compounds liste are formed by metals and non metals wich is an indication that they will have, at least, some ionic character.

Ergo, N and O will form covalent compounds, in this case NO2.

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Determine how much energy is needed to change 50 g of liquid water at 100°C to steam.

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Answer:

$Q=113,000J$

Explanation:

Hello there!

In this case, since the vaporization process is carried out in order to turn a liquid into a gas due to the addition of heat, we can use the following heat equation involving the heat of vaporization of water or any other substance:

$Q=m*\Delta _{vap}H$

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Which is positive due to the necessity of heat.

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2 years ago

Represente lo que ocurre mediante flechas cuando dos cargas negativas, dos cargas positivas y una carga positiva y otra negativa

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Answer:

Neutral.

Explicación:

Cuando dos cargas negativas, dos cargas positivas y una carga positiva y una carga negativa se unen, los átomos se vuelven neutrales porque las cargas opuestas cancelan el efecto de la otra. Si hay igual número de cargas y además son opuestas entre sí, entonces todas estas cargas cancelan el efecto de la otra formando el átomo neutral, pero si hay diferencia en el número de cargas, entonces la carga que es alta en número aparece en el átomos.

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An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i

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Answer:

$C_{metal}=126.6\frac{J}{g\°C}$

Explanation:

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In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

$Q_{metal}=-Q_{water}$

Which can be also written as:

$m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})$

Thus, since we need the specific heat of the metal, we solve for it as shown below:

$C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}$

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