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Murljashka [212]
3 years ago
10

Before free oxygen gas became abundant in the oceans and atmosphere, it reacted with a metal and produced what type of formation

s?
a. the ionosphere
b. the ozone layer
c. banded iron formations
d. stromatolites
Chemistry
2 answers:
Pavel [41]3 years ago
7 0
<h3><u>Answer;</u></h3>

Banded iron formations

Before free oxygen gas became abundant in the oceans and atmosphere, it reacted with a metal and produced <u>banded iron formations</u>.

<h3><u>Explanation</u>;</h3>
  • <u>Cyanobacteria are thought to have introduced oxygen into the early atmosphere.</u> This was important because oxygen played a key role in the evolution of eukaryotic organisms creating an ozone layer to shield life from UV rays.
  • Banded iron layers were formed in sea water as the result of oxygen released by the photosynthetic cyanobacteria.  As the photosynthetic organisms released oxygen, the available iron in the Earth's oceans precipitated out as iron oxides.
Vikentia [17]3 years ago
3 0
Before free oxygen gas became abundant in the oceans and atmosphere, it reacted with a metal and produced banded iron formations. 
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When we mix Cu wire with Silver Nitrate, the following reaction OCCURS.
kherson [118]

Answer:

Yes

2Ag + Cu(No3)2 -----(heat) ----- Cu + 2Agno3

When heat energy is passed through Cu(No3)2 then the reverse reaction will occur

3 0
3 years ago
What is the change in internal energy for each of the following situations? 1. q-7.9 J out of the system and w 3.6 J done on the
adoni [48]

Answer:

A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.

Explanation:

A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.

1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J

ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J

B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂

= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.

8 0
3 years ago
How many grams of NaOH are in a 225.0 mL sample of a 2.5 M solution of NaOH? (Molar mass of NaOH is 40.0 g/mol)
ladessa [460]

Answer

Approx. 4 g of NaOH.

Explanation:

Concentration

 

=

 

Moles

Volume of solution

Number of moles

 

=

 

Volume

×

concentration  

=

0.250

⋅

L

×

0.400

⋅

m

o

l

⋅

L

−

1

=

 

0.100

⋅

m

o

l

Mass of sodium hydroxide

 

=

 

0.100

⋅

m

o

l

×

40.0

⋅

g

⋅

m

o

l

−

1

=

?

?

⋅

g

Explanation:

6 0
3 years ago
CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.
kupik [55]

Answer:

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Explanation:

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The molar solubility of the solid cadmium fluoride = 0.0585 M

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NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0
3 years ago
Mention the main applications of analytical chemistry in geology.
Minchanka [31]

Answer:

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geo and chemistry

3 0
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