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Andru [333]
2 years ago
11

Where would the electrons in the 3d sublevel be found? Possibly in any shell only in the n = 3 energy level in energy levels &gt

; 3
Chemistry
1 answer:
postnew [5]2 years ago
4 0

Answer:

In the n = 3 energy level

Explanation:

There's is no further explanation for this.

All the electrons in an energy level are distribuited according to the period in the periodic table they are.

So, if we have an atom in period 1, like Hydrogen (H), that atom would only have 1 level energy (n = 1) and in that level, we only have the sub level 1s.

Electrons in the 3d sublevel, are found mostly in all the transition metals of period 3, and it can go from 1 to 10 electrons. To be with the 3d sub level it's neccesary that the energy level to be 3.

energy levels beyond that, like n = 4, we have electrons occupying the 3d sub level, so, primordly, the 3d is found only in energy level 3.

Hope this helps

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HELP PLEASE I DONT KNOW WHAT THIS MEANS IM CONFUSED
Zielflug [23.3K]

Explanation:

In an alpha emission, an alpha particle is ejected from a large nucleus. An alpha particle is simply a helium nucleus (atomic number is 2 and mass number is 4) so that means that the atomic number of the emitting nucleus decreases by 2 and its mass number decreases by 4.

5 0
3 years ago
When the mass of the boron atom is calculated, the mass of the electrons is ignored. Why is the mass of the electrons ignored?
Pavel [41]

Answer:

Because electron mass is so insignificant compared to proton and neutron mass that it can be ignored.

Explanation:

5 0
3 years ago
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Need to know how to do chemical reactions
melomori [17]
Chemical reactions happen by them self like if you hear baking soda and it makes a gas.
8 0
3 years ago
"An aqueous CaCl2 solution has a vapor pressure of 83.1mmHg at 50 ∘C. The vapor pressure of pure water at this temperature is 92
Lynna [10]

Answer : The the concentration of CaCl_2 in mass percent is, 41.18 %

Solution : Given,

Molar mass of water = 18 g/mole

Molar mass of CaCl_2 = 110.98 g/mole

First we have to calculate the mole fraction of solute.

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

\fac{p^o-p_s}{p^o}=X_B

where,

p^o = vapor pressure of the pure component (water) = 92.6 mmHg

p_s = vapor pressure of the solution = 83.1 mmHg

X_B = mole fraction of solute, (CaCl_2)

Now put all the given values in this formula, we get the mole fraction of solute.

\fac{92.6-83.1}{92.6}=X_B

X_B=0.102

Now we have to calculate the mole fraction of solvent (water).

As we know that,

X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898

The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.

Now we have to calculate the mass of solute, (CaCl_2) and solvent, (H_2O).

\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2

\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 11.32 + 16.164 = 27.484 g

Now we have to calculate the mass percent of CaCl_2

Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%

Therefore, the the concentration of CaCl_2 in mass percent is, 41.18 %

4 0
3 years ago
A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When he
grandymaker [24]

Answer:

The mass fraction of ferric oxide in the original sample :\frac{723}{3110}

Explanation:

Mass of the mixture = 3.110 g

Mass of Fe_2O_3=x

Mass of Al_2O_3=y

After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

Mass of mixture left after all the ferric oxide has reacted = 2.387 g

Mass of mixture left after all the ferric oxide has reacted = y

x=3.110 g- y=3.110 g - 2.387 g = 0.723 g

The mass fraction of ferric oxide in the original sample :

\frac{0.723 g}{3.110 g}=\frac{723}{3110}

5 0
3 years ago
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