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Flura [38]
3 years ago
12

A critical biotic factor in Yellowstone National Park is

Chemistry
1 answer:
dusya [7]3 years ago
5 0
The answer is E) proximity to water
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According to the theory of plate tectonics,
11Alexandr11 [23.1K]
A I’m sure of it because it only makes since one would think ✊
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3 years ago
Examine the photo of quartz below in which way does the quartz break
AveGali [126]
Can u add a link please 

4 0
3 years ago
What is the product of the unbalanced equation below?
Stolb23 [73]

Answer:

D

Explanation:

The answer is D. I'm not sure that it is a solid. I don't think it is a ppte, which is the only way it can be a true solid. It is ionic if the reaction is taking place in water and there is someway to start the reaction. Be that as it may, the internal balace numbers of the chemical produced is the only possible answer. The balanced eq;uatioon is

2Al + 3Br2 ==> 2AlBr3

3 0
2 years ago
Iron 3 oxide and carbon react to form iron and carbon dioxide. Balance the equation.
Oksi-84 [34.3K]

Answer:

2 Fe(iii)2O3 + 3 C ==> 2 Fe  + 3 CO2

Explanation:

First of all, you have to translate the words into an equation.

Fe(iii)2O3 + C ==> Fe  + CO2

The easiest way to tackle this is to start with the Oxygens and balance them. They must balance by going to the greatest common factor which is 6. So you multiply the molecule by whatever it takes to get the Oxygens to 6

2 Fe(iii)2O3 + C   ==>     Fe  + 3 CO2

Now work on the irons. There 2 on the left and just 1 on the right. So you need to multiply the iron by 2.

2 Fe(iii)2O3 + C ==> 2 Fe  + 3 CO2

Finally it is the turn of the carbons. There are 3 on the right, so you must make the carbon on the left = 3

2 Fe(iii)2O3 + 3 C ==> 2 Fe  + 3 CO2

And you are done.

5 0
3 years ago
Metallic copper is formed when aluminum reacts with copper(ii) sulfate. how many grams of metallic copper can be obtained when 5
exis [7]
Answer is: 127 grams <span>rams of metallic copper can be obtained.
</span>Balanced chemical reaction: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu.
m(Al) = 54.0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 54 g ÷ 27 g/mol.
n(Al) = 2 mol.
m(CuSO₄) = 319 g.
n(CuSO₄) = 319 g ÷ 159.6 g/mol.
n(CuSO₄) = 2 mol; limiting reactant.
From chemical reaction: n(CuSO₄) : n(Cu) = 3 : 3 (1 : 1).
n(Cu) = 2 mol.
n(Cu) = 2 mol · 63.55 g/mol.
n(Cu) = 127.1 g.
8 0
3 years ago
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