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3 years ago

What is the main type of weathering in deserts? weathering by organic acids chemical weathering physical weathering

Chemistry

1 answer:

Solnce55 [7]3 years ago

5 0

Answer:

chemical

Explanation:

Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.

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A 35 MGD surface Water treatement plant proposes to to use alum Coagulation for turbidity reduction using 44 mg/ L alum. the raw

CaHeK987 [17]

C) is the water alkainty adequate. I.e

5 0

2 years ago

A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch

ivanzaharov [21]

Answer:

The reactions free energy $\Delta G = -49.36 kJ$

Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

The pressure of (Cl) gas is $P_{Cl} = 9.15 \ atm$

The pressure of nitrosly chloride (NOCl) is $P_{(NOCl)} = 7.70 \ atm$

The reaction is

$2NO_{(g)} + Cl_2 (g)$ ⇆ $2 NOCl_{(g)}$

From the reaction we can mathematically evaluate the $\Delta G^o$ (Standard state free energy ) as

$\Delta G^o = 2 \Delta G^o _{NOCl} - \Delta G^o _{Cl_2} - 2 \Delta G^o _{NO}$

The Standard state free energy for NO is constant with a value

$\Delta G^o _{NO} = 86.55 kJ/mol$

The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

The Standard state free energy for $NOCl$ is constant with a value

$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

$= -43 kJ/mol$

The pressure constant is evaluated as

$Q = \frac{Pressure \ of \ product }{ Pressure \ of \ reactant }$

Substituting values

$Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}$

$= 0.0765$

The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

T is temperature in K with a given value of $T = 25+273 = 298 K$

Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

$= -43-6.36$

$\Delta G = -49.36 kJ$

4 0

3 years ago

15.00 g of NH4HS(s) is introduced into a 500. mL flask at 25 °C, the flask is sealed, and the system is allowed to reach equilib

Ahat [919]

Answer:

0.328 atm

Explanation:

Kp is the equilibrium constant calculated based on the pressure, and it depends only on the gas substances. It will be the multiplication of partial pressures of the products raised to their coefficients divided by the multiplication of partial pressures of the reactants raised to their coefficients.

For the equation given, the stoichiometry is 1 mol of NH₃ for 1 mol of H₂S, so they will have the same partial pressure in equilibrium, let's call it p. So:

Kp = pxp

0.108 = p²

p = √0.108

p = 0.328 atm, which is the partial pressure of the ammonia.

3 0

3 years ago

what is the pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4

Anna71 [15]

Answer:

Explanation:

First, find the hydronium ion concentration of the solution with a pH of 4.

[H₃O⁺] = 10^-pH

[H₃O⁺] = 10⁻⁴

[H₃O⁺] = 1 × 10⁻⁴

Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.

[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01

Lastly, find the pH.

pH = -log [H₃O⁺]

pH = -log (0.01)

pH = 2

The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.

Hope this helps.

4 0

3 years ago

Formal charge of Cl in CHCl3

Vsevolod [243]

FC(C) = 4 - 0.5*8 - 0 = 0) FC(Cl) = 7 - 0.5*2 - 6 = 0

6 0

3 years ago