Características de la báscula granataria porfi

How have ideas about the atom changed in the last two centureies?

dsp73

In the past a scientist named dalton produced an atomic theory. There were certain problems regarding his views. So, later on scientists like chadwick, rutherford and thompson added some fresh light for the real identification of the atom.
dalton said the atom was the smallest unit and it CANNOT BE DIVIDED ANY FURTHER.
NOTE... this was renected with the discovery of the proton, neutron and electron as the sub atomic particles.

8 0

3 years ago

Manganese, Mn, forms two ions, one with a 2+ charge and one with a 3+ charge. What is the formula for manganese (II) sulfide?

maria [59]

It would be MnSO4

The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge

These will cancel out making it plain MnSO4

If it was manganese (iii) sulfide the answer would be Mn2(SO4)3

4 0

3 years ago

Consider the elements: Na, Mg, Al, Si, P.

Olegator [25]

Answer:

The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.

a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.

The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.

b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.

c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.

d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,

1s2 2s2 2p6 3s2

In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.

8 0

3 years ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$