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Anastaziya [24]
3 years ago
15

Explain why it is not a good idea to throw an aerosol can into fire. Which gas law applies?

Chemistry
2 answers:
frosja888 [35]3 years ago
7 0

Answer:

This is because the pressure will build so up so much that the can will burst (and explode if there is a naked flame nearby) the gas pressure increases. more of the liquefied propellant turns into a gas.  

This is Boyle's Law. It states pressure and volume have an inverse relationship.

Explanation:

suter [353]3 years ago
4 0

Answer:

The gas inside the can and the can’s volume are both constant.

The gas pressure increases with increasing temperature.

The can will burst if the pressure becomes great enough.

The gas law that applies is Gay-Lussac’s law.

Explanation:

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This is a solution whose solvent is water.
Anastasy [175]
It would be the Aqueous solution
8 0
3 years ago
What are the limitations of using chemical indicators in determining the pH of a given solution?
tino4ka555 [31]

Answer:

1) chemical indicators won't work above it's pH range so therefore it probably won't change colour.

2) the solution should be clear and colourless to see colour change.

3) indicators tend to be of low accuracy so it's not 100% reliable.

5 0
3 years ago
Calculate the mass of butane needed to produce 97.4 g of carbon dioxide. Express your answer to three significant figures and in
Vsevolod [243]

Answer:

32.1 g

Explanation:

Step 1: Write the balanced combustion reaction

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 97.4 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

97.4 g × 1 mol/44.01 g = 2.21 mol

Step 3: Calculate the moles of butane that produced 2.21 moles of carbon dioxide

The molar ratio of C₄H₁₀ to CO₂ is 1:4. The moles of C₄H₁₀ required are 1/4 × 2.21 mol = 0.553 mol

Step 4: Calculate the mass corresponding to 0.553 moles of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

0.553 mol × 58.12 g/mol = 32.1 g

4 0
3 years ago
What property do atoms of these elements have that helps make the molecules polar
DENIUS [597]
You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
even distribution -------> non-polar
4 0
3 years ago
Given the following information:Br2 bond energy = 193 kJ/mol F2 bond energy = 154 kJ/mol 1/2Br2(g) + 3/2F2(g) -> BrF3(g) = –3
arlik [135]

Answer:

C) 712 KJ/mol

Explanation:

  • ΔH°r = Σ Eb broken - Σ Eb formed
  • 1/2Br2(g) + 3/2F2(g) → BrF3(g)

∴ ΔH°r = - 384 KJ/mol

∴ Br2 Eb = 193 KJ/mol

∴ F2 Eb = 154 KJ/mol

⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)

⇒ Σ Eb broken =  (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol

∴ Eb formed: Br-F

⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r

⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )

⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol

8 0
3 years ago
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