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beks73 [17]
3 years ago
8

Rain washing away soil from a hillside

Chemistry
1 answer:
Aneli [31]3 years ago
8 0

Answer:

deposition

Explanation:

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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
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