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3 years ago

How many electrons are in the highest occupied energy level of an element in group 5a

Chemistry

1 answer:

Setler79 [48]3 years ago

5 0

Answer:

3 electrons

Explanation:

The elements of group 5A of the periodic table are the following:

nitrogen, sodium, arsenic, antimony and bismuth.

To determine the amount of electrons in the last energy level we need to know the electronic configuration of each element.

Electronic configuration

N: [He] 2s2 2p3

P: 1s2 2s2 2p6 3s2 3p3

As: [Ar] 3d10 4s2 4p3

Sb: [Kr] 4d10 5s2 5p3

Bi: [Xe] 4f14 5d10 6s2 6p3

As we can see they all have 3 electrons in their last energy level .

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find the total pressure of a gas that initially occupied 27 L at 32 degrees Celsius and 2.5 atm, if the final conditions are 12

yawa3891 [41]

Answer: Total pressure of the gas will be 0.716atm.

Explanation: We are given a gas having initial conditions as

V = 27L

T = 32°C = 305K

P = 2.5atm

As the gas remains same, number of moles of a gas will also be same for initial and final conditions. To calculate the number of moles, we use ideal gas equation, which is,

$PV=nRT$ .......(1)

where, R = gas constant = $\text{0.08206 L atm }mol^{-1} K^{-1}$

For calculating number of moles:

$n=\frac{PV}{RT}$

Putting the values of initial condition in this equation, we get

$n=\frac{(2.5atm)(27L)}{\text{(0.08206 L atm }mol^{-1} K^{-1})(305K)}$

n = 2.696 mol

Now, the final conditions are,

V = 88.0L

T = 12°C = 285K

n = 2.696 mol (calculated above)

P = ? atm

Again using equation 1, we get

$P=\frac{nRT}{V}$

$P=\frac{(2.696mol)(\text{0.08206 L atm }mol^{-1} K^{-1})(285K)}{88.0L}$

P = 0.716atm.

7 0

3 years ago

Read 2 more answers

Consider the equilibrium

vladimir1956 [14]

Answer:

$Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}$

$\Delta _rG=1.01x10^5J/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)$

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

$p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar$

$Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141$

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

$Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}$

Finally, the Gibbs free energy for the reaction at 298.15K is:

$\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol$

Best regards.

3 0

4 years ago

• Calculate the entropy change when 1.00 mol of

Musya8 [376]

Answer:

the enthalpy change of this is 5.02mol

3 0

2 years ago

Boron has primarily two isotopes, one with an atomic mass of 11.0 amu and another with an atomic mass of 10.0 amu. If the abunda

Masja [62]

Answer:

The atomic mass of the boron atom would be 10.135

Explanation:

This is generally known as relative atomic mass.

Relative atomic mass or atomic weight is a physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass of 1/12 of the mass of a carbon-12 atom. Since both quantities in the ratio are masses, the resulting value is dimensionless; hence the value is said to be relative and does not have a unit.

Note that the relative atomic mass of atoms is not always a whole number because of it being isotopic in nature.

Divide each abundance by 100 then multiply by atomic mass
Do that for each isotope, then add the two result. Thus

Relative atomic mass of Boron = (18.5/100 x 11) + (81/100 x 10)

= 2.035 + 8.1

= 10.135

5 0

3 years ago

Compound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when

DochEvi [55]

Answer:

2,2,3,3-tetrapropyloxirane

Explanation:

In this case, we have to know first the alkene that will react with the peroxyacid. So:

What do we know about the unknown alkene?

We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If $(CH_3CH_2CH_2)_2C=O$ is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.

What is the product with the peroxyacid?

This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)

3 0

3 years ago