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xz_007 [3.2K]
3 years ago
14

What are the three types of combustion reactions

Chemistry
2 answers:
Jlenok [28]3 years ago
7 0

Answer:

Three types are: Rapid Combustion, Complete Combustion, and Spontaneous Combustion.

Explanation:

Note: there are more types! This is just three random ones I picked to list. Hope this helps! :)

SVEN [57.7K]3 years ago
4 0

Answer:

Slow combustion

Spontaneous combustion

Explosive combustion

Explanation:

-Slow combustion reactions: Occurs at low temperatures. Cellular respiration is an example.

-Spontaneous combustion reactions: Occurs suddenly without an outside heat source. The heat source is the result of oxidation.

-Explosive combustion reactions: Involves an oxidizing agent.

hopefully this helped :3

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Convert 22.4 kg/L to kg/mL
liberstina [14]
1 kg/L -------------- 0.001 kg/mL
22.4 kg/L --------- ??

22.4 x 0.001 / 1 => 0.0224 kg/mL
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3 years ago
Akio just got over having the flu. His brother just got the flu. His parents are not worried about Akio getting it again because
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Answer:

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If a recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb what volume in liters
Bad White [126]

The volume in liters of 3245 aluminum cans is 24.8 L.

A recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb. The mass of 3245 aluminum cans is:

3245 cans \times \frac{1lb}{22cans} = 147.5 lb

To convert mass to volume, we need the density of aluminum (5.95 lb/L). The volume corresponding to 147.5 lb of aluminum is:

147.5 lb \times \frac{1L}{5.95 lb} = 24.8 L

The volume in liters of 3245 aluminum cans is 24.8 L.

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2 years ago
Bacteria cells are different from plant and animal cells because they don't have what???!
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Plants and animals are multi-cellular organisms composed of eukaryotic cells, while bacteria are single-cell prokaryotic organisms. Each eukaryotic cell of a plant or animal includes a central nucleus containing DNA and membrane-bound organelles, such as endoplasmic reticulum and mitochondria. A bacterial cell has no nucleus or membrane-bound organelles.

hope it helps! :)

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3 0
3 years ago
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An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas
n200080 [17]

Answer:

  • The limiting reactant is lead(II) nitrate.
  • 7.20 g of precipitate are formed.
  • 1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

  • Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:

  • 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
  • 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:

  • 0.0296 mol Pb(NO₃)₂ \frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}} * \frac{278.1g}{1molPbCl_{2}} * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

  • 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:

  • 0.0259 mol Pb(NO₃)₂ \frac{2molKCl}{1molPb(NO_{3})_2} * \frac{74.55g}{1molKCl} = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

  • 5.76 - 3.86 = 1.9 g KCl

8 0
3 years ago
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