No it depends on the molecules strength
<span><u><em>Answer:</em></u>
combustion reaction
<u><em>Explanation:</em></u>
In chemistry, a <u>combustion reaction</u> is defined as a reaction between an oxidant and any compound that leads to the production of another compound along with a huge amount of heat.
<u>Now, let's check the reaction given:</u>
C</span>₃H₈<span> + 5 O</span>₂<span> --> 3 CO</span>₂<span> + 4 H</span>₂O<span>
<u>The oxidant</u> is oxygen gas
<u>The compound reacting</u> is propane
<u>The compound produced</u> is carbon dioxide along with water vapor and heat
Therefore, the given reaction is a combustion reaction
Hope this helps :)</span>
It is called DISPLACEMENT of the object....
Answer:
solubility of X in water at 17.0 is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = = 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>