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What is the correct orbital diagram for carbon

Viefleur [7K]

This is a basic orbital diagram for carbon

5 0

3 years ago

Which of the following is true?

Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the <u>energy available to do work</u>, of a system for a constant-temperature process is:

$ΔG = ΔH - TΔS$

When ΔG < 0 the reaction is spontaneous in the forward direction.

When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

When ΔG = 0 the system is at equilibrium.

If <u>both ΔH and ΔS are positive</u>, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.

3 0

3 years ago

What volume would 56.2 mL of gas at 820 mm of Hg occupy at 720 mm of Hg?

Andreyy89

Answer:

49.35 mL

Explanation:

Given: 56.2 mL of gas

To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg

Solution:

At 820 mm of Hg, volume of gas is 56.2 mL

At 1 mm of Hg, volume of gas is $\frac{56.2}{820}$

At 720 mm of Hg, volume of gas is $\frac{56.2}{820}(720)=49.35\,\,mL$

3 0

3 years ago

Calculate the standard free-energy change at 25 ∘C for the following reaction:

lianna [129]

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$ ( $\Delta G^{0}$ ): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

8 0

3 years ago

Read 2 more answers

8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]

yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡ ΔH $_{f}$ = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8 kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5 kJ/mol

Equation 4 ≡ BE(Cl2) = 226 kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332 kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

4 0

3 years ago