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kenny6666 [7]
3 years ago
15

Why wasn't oxidation type chemical weathering common more than 2 billion years ago?

Chemistry
1 answer:
scZoUnD [109]3 years ago
6 0

Answer:

            Due to deficiency of Oxygen in atmosphere.

Explanation:

                   Oxidation is defined in two ways, 1) The addition of Oxygen and Removal of Hydrogen 2) The removal of electrons.

                   We will discuss the first definition for this question. The addition of oxygen to various compounds results in the oxidation of that compound.

Example: Oxidation of Methane;

                                   CH₄  +  2 O₂    →    CO₂  +  2 H₂O

In this example oxygen is added to methane and hydrogen is being removed from methane.

Oxidation type chemical weathering common more than 2 billion years ago because of less amount of Oxygen in atmosphere and the oxygen which was produced by plants reacted with different metals like Iron forming precipitates which precipitated to the bottom of oceans.

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Answer:

Endothermic Reaction

Explanation:

5 0
3 years ago
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If a 0.4681 g Mg strip reacts with 0.650 M HCl in a 139.3 mL flask at 25oC, what is the minimum volume (mL) of HCl needed to com
Marizza181 [45]
The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,

    Mg + 2HCl --> H2 + MgCl2

The number of moles of HCl that is needed for the reaction is calculated below.
    n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
    n = 0.0385 mols HCl

From the given concentration, we calculate for the required volume. 
    V = 0.0385 mols HCl/(0.650 mols/L)
     V = 0.05926 L or 59.26 mL

<em>Answer: 59.26 mL of HCl</em>
7 0
3 years ago
How many moles of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an excess of lead
frez [133]
The answer would be 313 g
4 0
3 years ago
a) Provide equation of K of this reaction, use symbol " ^ " for exponents. That means 1000 = 10^3 and 1/100 is 10^(-2). b) How m
stealth61 [152]

Answer and Explanation:

a. The equation of K of this reaction is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

K = \frac{(D)^5 (E)^7 (F)}{(A)^3 (B)^5 (C)^4}

b. The moles of compound F is shown below:-

3 A + 5 B + 4 C↔5 D + 7 E + F

               2 moles

Now, the mole of produced is

= \frac{1}{4} \times \ moles\ of\ c

Now, we will the value of c by using the above equation

= \frac{1}{4} \times 2

After solving the above equation we will get

0.5 moles

5 0
3 years ago
In a unimolecular reaction with twice as much starting material as product at equilibrium, what is the value of Keq? Is ΔG o pos
lubasha [3.4K]

Answer : The value of K_{eq} is, 0.5 and \Delta G= positive.

Explanation :

The unimolecular reaction is:

A\rightarrow B

In unimolecular reaction, the starting material is 2 times to the product.

A=2B      .........(1)

As we know that:

K_{eq}=\frac{B}{A}     ...........(2)

Now substitute equation 1 in 2, we get:

K_{eq}=\frac{\frac{A}{2}}{A}

K_{eq}=0.5

Now we have to calculate the value of \Delta G^o at 298 K.

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

K_{eq} = equilibrium constant = 0.5

Now put all the given values on the above formula, we get:

\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.5)

\Delta G^o=1717.32J/mol

Thus, the value of \Delta G^o at 298 K is, 1717.32 J/mol

As we know that:

\Delta G= +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

Thus, the \Delta G= +ve. So, the reaction is non spontaneous.

8 0
3 years ago
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