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KIM [24]
3 years ago
7

What does C=o represent

Chemistry
1 answer:
Novay_Z [31]3 years ago
7 0

Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is slightly less dense than air. ... Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.  


I think it's probably not right.

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What is the answers? <br> Is my answer right?
V125BC [204]

Answer:

I say the correct answers are primary and secondary and teriary.

Explanation:

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5 0
3 years ago
Four students presented different analogies to describe the formation of an ionic bond.Student A: A tug of war between seven mid
MissTica

Answer:

The answer is A

Explanation:

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3 years ago
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.

8 0
3 years ago
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umka21 [38]

Answer: Bacteria✅

Explanation:

4 0
2 years ago
In a disproportionation reaction, the disproportionate substance
trapecia [35]

Answer:

The answer is C: has at least three oxidation states.

Explanation:

you're welcome

7 0
3 years ago
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