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HACTEHA [7]
3 years ago
5

The five boxes below represent a geologic cross-section with five layers that represent the rocks as they would from when earth'

s atmosphere was developing (A is on top and E is on the bottom). Label the cross section with the appropriate rock from the word bank below. Then, explain your answer in relation to the history of the earth's atmosphere.
Chemistry
1 answer:
quester [9]3 years ago
7 0

Answer:

A Limestone Layers

B Banded Iron formations

C Metaphoric rocks with fine grains  

D Red Beds

E Chert Layers

Explanation:

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Which statement is correct about charges?
V125BC [204]

Answer:

  • A.Opposite charges repel
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3 years ago
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You mix 265.0 mL of 1.20 M lead(II) nitrate with 293 mL of 1.55 M potassium iodide. The lead(II) iodide is insoluble. What amoun
slava [35]

Answer:

105 grams PbI₂

Explanation:

Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)

moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole

moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant

moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂

grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)

7 0
3 years ago
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Two plates slide past each other in opposite directions at a ____ boundary.
Kitty [74]

Answer: plate

Explanation: Plates grinding past each other in opposite directions create faults called transform faults. Powerful earthquakes often strike along these boundaries. The San Andreas Fault is a transform plate boundary that separates the North American Plate from the Pacific Plate.

4 0
3 years ago
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Any suggestions or answers
Eva8 [605]

Answer: correct answer is D

It can be traced example in copper rod or iron or any other pure metal

3 0
3 years ago
Given the equation:
Citrus2011 [14]

Answer: a)  Fe is the limiting reagent

b) 3.59 g

c) 11.6g

Explanation:

4Al_2O_3+9Fe\rightarrow 3Fe_3O_4+8Al  

To calculate the moles :

   

\text{Moles of} Al_2O_3=\frac{27.5g}{102g/mol}=0.27moles

\text{Moles of} Fe=\frac{8.4g}{56g/mol}=0.15moles

According to stoichiometry :

a) 9 moles of Fe  require= 4 moles of Al_2O_3

Thus 0.15 moles of Fe will require=\frac{4}{9}\times 0.15=0.067moles  of Al

Thus Fe is the limiting reagent as it limits the formation of product and Al is the excess reagent.

b) As 9 moles of Fe give = 8 moles of Al

Thus 0.15 moles of Fe give =\frac{8}{9}\times 0.15=0.133moles  of Al

Mass of Al=moles\times {\text {Molar mass}}=0.133moles\times 27g/mol=3.59g

c) As 9 moles of Fe give = 3 moles of Fe_3O_4

Thus 0.15 moles of Fe give =\frac{3}{9}\times 0.15=0.05moles  of Fe_3O_4

Mass of Fe_3O_4=moles\times {\text {Molar mass}}=0..05moles\times 231.5g/mol=11.6g

8 0
3 years ago
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