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Use the periodic table to determine the electron configuration for Ca and Pm in noble-gas notation Ca: [Ar]4s2 [Ar]4s1 [Ar]3s2 [

sveta [45]

Answer:

1) Ca: [Ar]4s²
2) Pm: [Xe]6s²4f⁵

Explanation:

1) Ca:

Its atomic number is 20. So it has 20 protons and 20 electrons.

Since it is in the row (period) 4 the noble gas before it is Ar, and the electron configuration is that of Argon whose atomic number is 18.

So, you have two more electrons (20 - 18 = 2) to distribute.

Those two electrons go the the orbital 4s.

Finally, the electron configuration is [Ar] 4s².

2) Pm

The atomic number of Pm is 61, so it has 61 protons and 61 electrons.

Pm is in the row (period) 6. So, the noble gas before Pm is Xe.

The atomic number of Xe is 54.

Therefore, you have to distribute 61 - 54 = 7 electrons on the orbitals 6s and 4f.

The resultant distribution for Pm is: [Xe]6s² 4f⁵.

5 0

2 years ago

Read 2 more answers

What is the answers?

ra1l [238]

the numbers are going to be small so like a power but its at the bottom

NH3, H2O2, NHO2

5 0

3 years ago

Is this an aromatic hydrocarbon? Why or why not?

swat32

The correct answer is D

3 0

3 years ago

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Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly

trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of $Ag_2O$ = 25.0 g

Mass of $C_{10}H_{10}N_4SO_2$ = 50.0 g

Molar mass of $Ag_2O$ = 231.7 g/mole

Molar mass of $C_{10}H_{10}N_4SO_2$ = 250.3 g/mole

Molar mass of $AgC_{10}H_{9}N_4SO_2$ = 357.1 g/mole

First we have to calculate the moles of $Ag_2O$ and $C_{10}H_{10}N_4SO_2$ .

$\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles$

$\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 1 mole of $Ag_2O$

So, 0.1998 moles of $C_{10}H_{10}N_4SO_2$ react with $\frac{0.1998}{2}=0.0999$ moles of $Ag_2O$

From this we conclude that, $Ag_2O$ is an excess reagent because the given moles are greater than the required moles and $C_{10}H_{10}N_4SO_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgC_{10}H_9N_4SO_2$

From the reaction, we conclude that

As, 2 mole of $C_{10}H_{10}N_4SO_2$ react with 2 mole of $AgC_{10}H_9N_4SO_2$

So, 0.1998 mole of $C_{10}H_{10}N_4SO_2$ react with 0.1998 mole of $AgC_{10}H_9N_4SO_2$

Now we have to calculate the mass of $AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2$

$\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g$

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0

3 years ago

Calculate the total binding energy, and the binding energy per nucleon, for 73Li. The masses of the atoms of 73Li and 11H are 7.

charle [14.2K]

Answer:

5.60 Mev

Explanation:

Binding Energy is termed as the total energy required for the breakdown of a nucleus to its component nucleons

From the given question;

There are 3H and 4n,

Therefore;

since the masses of the atoms are 7.016004 u and 1.007825 u,

Then

3(1.007825) + 4(1.008665) = 7.058135 u

( 7.058135 - 7.016004) × (931.5) = 39.24 MeV total biding energy.

The energy per nucleon is therefore calculate as:

$\frac {39.24 \ \ MeV }{7}$

=5.60 Mev

Thus, the total binding energy is = 5.60 Mev

8 0

2 years ago