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Roman55 [17]
2 years ago
10

Use the successive ionization energies for this unknown element to identify the family it belongs to.

Chemistry
1 answer:
NNADVOKAT [17]2 years ago
4 0

Answer: Belongs to the group 2A

Explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low  ionization energies  for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

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A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist
REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

Temperature of the water bath before the reaction = 25 °C

Temperature of the water bath after the completion of the reaction = 28 °C

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

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As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

3 0
3 years ago
For the balanced equation shown below, how many moles of o2 will react with 0.3020 moles of co2? 2c2h5oh + 6o2 → 4co2 + 6h2o que
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The balanced reaction for combustion is as follows ;
2C₂H₅OH + 6O₂ ---> 4CO₂ + 6H₂O
the stoichiometry of C₂H₅OH to O₂ is 2:6
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when 1 mol of C₂H₅OH reacts with 6/2 mol of O₂,
then 0.3020 mol of C₂H₅OH reacts with - 6/2 x 0.3020
therefore number of O₂ moles reacted = 0.91 mol
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