I have to have the one in the front door open at the door and I’m sorry I don’t know how much you want me to do it but I’m not trying to be rude but I’m not going on a trip and I have a friend that I’m not trying to be rude but I just hate to say that I’m sorry I’m
Answer:
answer is C
Explanation:
encourage the release of carbon dioxide from the stems
Answer:
No, you can not calculate the solubility of X in water at 17 0C.
Explanation:
Solubility refers to the amount of a substance that dissolves in 1000 L of water.
To calculate the solubility of a solute in water, all the water is evaporated and the solid is carefully collected, washed, dried and weighed. The mass of solid obtained can now be used to calculate the solubility of the solute in water as long as there was no loss in mass of solid during the experiment.
In this case, the student threw away part of the solid that precipitated. As a result of this, the mass of solid obtained at the end of the experiment is not exactly the total mass of solute that dissolved in the solvent. Hence, the solubility of X in water at 17 0C can not be accurately calculated.
Answer:
- <u><em>The volume of CO₂(g) produced at STP when 0.05 moles of C₂H₄(g) was burnt in O₂(g) is 2.24dm</em></u><em><u>³</u></em>
Explanation:
The question is incomplete.
This is the complete question:
<em>Consider the reaction by the following equation:</em>
<em />
<em> C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)</em>
<em />
<em>The volume of CO₂(g) produced at s.t.p when 0.05 moles of C₂H₄(g) was burnt in O₂(g) is ________</em>
<em />
<em>[Molar Volume of gas = 22.4dm³]</em>
<em />
<em> A. 1.12dm³</em>
<em> B. 2.24dm³</em>
<em> C. 3.72dm³</em>
<em> D. 4.48dm³</em>
<h2>Solution</h2>
<u />
<u>1. Write the mole ratio between CO₂(g) and C₂H₄(g)</u>
- 1 mol C₂H₄(g) : 2 mol CO₂(g)
<u>2. Multiply the 0.05 moles of C₂H₄(g) by the mole ratio</u>
- 0.05 mol C₂H₄ × 2 mol CO₂ / 1 mol C₂H₄(g) = 0.10 mol CO₂
<u>3. Convert moles of CO₂ to volume at STP using the molar volume at STP</u>
- 0.10 mol CO₂ × 22.4 dm³ / mol = 2.24 dm³