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3 years ago

Describe the spin of carbon-14's valence electrons and explain how you came about your answer.

Chemistry

1 answer:

Snezhnost [94]3 years ago

8 0

we have to know the spin of valence electrons of carbon-14

There are four unpaired electron which are called as valence electron also.The spin of the four unpaired electron is either upfilled or down filled.

The ground state electronic configuration of C-atom is 1s²2s²2p² and one electron from 2s orbital gets excited to 2p orbital. The elctronic configuration in excited state is 1s²2s¹ $2p_{x}^{1} 2p_{y}^{1}2p_{z}^{1}$ .

The electron jumps because half-filled orbitals are more stable. Exchange energy is less than pairing energy.

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The actions described in the passage are best understood in the context of which of the following?

soldi70 [24.7K]

C
I have had this question on a test before!! Hope this helps

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3 years ago

Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu

MAVERICK [17]

Answer : The initial temperature of system 2 is, $19.415^oC$

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

$-q_1=q_2$

$m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)$

The mass remains same.

where,

$C_1$ = heat capacity of system 1 = 19.9 J/mole.K

$C_2$ = heat capacity of system 2 = 28.2 J/mole.K

$T_f$ = final temperature of system = $30^oC=273+30=303K$

$T_1$ = initial temperature of system 1 = $45^oC=273+45=318K$

$T_2$ = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

$-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K$

$T_2=292.415K$

$T_2=292.415-273=19.415^oC$

Therefore, the initial temperature of system 2 is, $19.415^oC$

8 0

4 years ago

The density of an object with a volume of 5.0mL and a mass of 20.0g is ____________.

Lina20 [59]

whenn you have sublimation

6 0

3 years ago

Read 2 more answers

A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the

likoan [24]

Answer:

The final temperature of the solution is 44.8 °C

Explanation:

assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:

Q dis + Q sol = 0

Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis= -83.3 KJ/mol . And the molecular weight is

M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol

Q dis = q dis * n = q dis * m/M = -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ

Qdis= -9.84 KJ

Also Qsol = ms * Cs * (T - Ti)

therefore

ms * Cs * (T - Ti) + Qdis = 0

T= Ti - Qdis * (ms * Cs )^-1 =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ

T= 44.8 °C

7 0

4 years ago

A mixture of co(g) and h2(g) is pumped into a previously evacuated 2l reaction vessel the total pressure of the reaction system

swat32

Answer is: total pressure of the system is 2.4 atm.

Boyle's Law (the pressure volume law): volume of a given amount of gas held varies inversely with the applied pressure when the temperature and mass are constant.

p₁V₁ = p₂V₂ (the product of the initial volume and pressure is equal to the product of the volume and pressure after a change).

1.2 atm · 2 L = p₂ · 1 L.

p₂ = 1.2 atm · 2 L / 1 L.

p₂ = 2.4 atm.

When pressure goes up, volume goes down.

When volume goes up, pressure goes down.

7 0

3 years ago

Read 2 more answers