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docker41 [41]
2 years ago
14

Amanda paired with Janice while Deja paired with Eden for a chemistry project. For the Earth science project, Amanda paired with

Eden while Deja paired with Janice. Which type of chemical reaction does the situation demonstrate?
Decomposition reaction
Single replacement
Synthesis reaction
Double replacement reaction
Chemistry
2 answers:
Black_prince [1.1K]2 years ago
4 0

Answer:

A

Explanation:

pashok25 [27]2 years ago
3 0

The chemical reaction that the situation demonstrates would be a double replacement reaction.

In double replacement reactions, the two reactants participating in the reaction are similarly built in terms of their chemical bonds and they exchange ions to form the products of the reaction. Two products are also formed from the two reactants.

It is as opposed to single replacement reactions in which the two reactants are not similar bond-wise. One of the reactants replaces or displaces one of the ions in another reactant.

In this case, the situation can be represented as follows:

Amanda-Janice + Deja-Eden ----> Amanda-Eden + Deja-Janice

Thus, it is a form of double replacement reaction.

More on double replacement reactions can be found here: brainly.com/question/392491?referrer=searchResults

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What are some types of chemical weathering?​
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Read 2 more answers
Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

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Explanation:

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Answer:

might help

Explanation:

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