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finlep [7]
4 years ago
6

how the problem of removing excess carbon dioxide from the atmosphere is connected to the law of conversation of matter

Chemistry
2 answers:
Neko [114]4 years ago
7 0

Answer:

According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and type of atoms after the chemical change as were present before the chemical change.

This same principle can be applied to removing excess CO2 in the atmosphere.

Hope this helps...Mark as Brainliest pls!!

USPshnik [31]4 years ago
5 0

Possible Answer: The problem of removing excess carbon dioxide from the atmosphere is related to the law of conservation of matter because the law states that matter cannot be destroyed. This means, it is not a simple fix because the matter that is the excess carbon dioxide cannot just be destroyed. Instead, it must be (a long with future emissions of the gas causing an excess) redistributed in a way that doesn't cause harm.

Explanation of the law: law of conservation of matter: matter is neither created nor destroyed. Example, Google says, ". . . so we must have the same number and type of atoms after the chemical change as were present before the chemical change."

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Answer:

Explanation:

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5 0
3 years ago
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The total pressure in a tank fill with a mixture of gasses : Oxygen, Helium, and Argon is 15.3 atm. The phe = 3.2 atm , Po= 7.4
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Answer:

Partial pressure of Ar = 4.7 atm

Explanation:

Total pressure of a mixture of gases = Sum of partial pressure of each gas

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Total pressure = Partial pressure O₂ + Partial pressure of He + Partial pressure of Ar

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3 years ago
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Answer:

2.82 g

Explanation:

Step 1: Write the balanced precipitation reaction

3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ⇒ 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)

Step 2: Calculate the reacting moles of Ba(NO₃)₂

45.0 mL (0.0450 L) of 0.548 M Ba(NO₃)₂ react.

0.0450 L × 0.548 mol/L = 0.0247 mol

Step 3: Calculate the moles of Al₂(SO₄)₃ that react with 0.0247 moles of Ba(NO₃)₂

The molar ratio of Ba(NO₃)₂ to Al₂(SO₄)₃ is 3:1. The reacting moles of Al₂(SO₄)₃ are 1/3 × 0.0247 mol = 8.23 × 10⁻³ mol

Step 4: Calculate the mass corresponding to 8.23 × 10⁻³ moles of Al₂(SO₄)₃

The molar mass of Al₂(SO₄)₃ is 342.2 g/mol.

8.23 × 10⁻³ mol × 342.2 g/mol = 2.82 g

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Answer:

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