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ikadub [295]
3 years ago
7

A green box contains a blue box and a pink box. The blue box contains C H subscript 4 + 2 O subscript 2. An arrow then leads in

the green box to the pink box, which says C O subscript 2 + 2 H subscript 2 O. Lines lead from labels to parts of the boxes. Label A leads to the beginning of the green box. Label B leads to the + in the blue box. Label C leads to the arrow. Label D leads to the 2 in 2 H subscript 2 O. Label E leads to the end of the pink box.
Identify each part of this chemical equation that describes the burning of methane and oxygen.

B (blue box):

D (number):

E (purple box):
Chemistry
2 answers:
Artemon [7]3 years ago
6 0

Answer:

The Correct Answers Are: The Entire Green Box: Chemical Equation B (The Blue Box): Reactants C (The Arrow): Reacts To Forms D (The Numbers): Coefficient E (The Purple Box): products

blsea [12.9K]3 years ago
4 0

Answer:

Identify each part of this chemical equation, which describes the breakdown of water into hydrogen and oxygen.

A (the entire green box):

✔ chemical equation

B (the blue box):

✔ reactants

C (the arrow):

✔ reacts to form

D (the purple box):

✔ products

Explanation:

I did it on edg and got it right

also can I have some yummy brainiest  

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Explanation:

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Explain in YOUR words what factors determine which type of fossil fuel forms; coal, oil, or gas.
MariettaO [177]

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Explanation:

8 0
3 years ago
How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?
hoa [83]

Answer:

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

______

NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g.  Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.  

8 0
3 years ago
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