3. A magnet attracts a piece of metal. How is this event like a chemical bond?

Which group 4A element has the highest ionization energy?

Gelneren [198K]

The ionizatin energy is the is the energy required to remove one electron from from a gaseous atom.

There is a clear trend for the first ionization energies (the ionization energy to remove the first electron) of the atoms.

Inside a group (a column of the periodic table) the ionization energy increases downwards .

Then, the element that has the greatest atomic number in the group has the highest ionization energy (it is more difficult to form its ion).

As per this trend, in the group 4A lead, Pb (atomic number 82) is the natural element with the highest ionization energy. But there is one artificial element in this group, whose atomic number is 114. It is called Flerovium and as per the rule it shall have a higher ionization energy.

4 0

3 years ago

Which of the following acts as a catalyst in catalytic converters?

yaroslaw [1]

Answer: B. Metal

Explanation: The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium Hope This Helped (=^･-･^=)

4 0

3 years ago

Read 2 more answers

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$