Answer:
Haven't evaporated all of the water
Explanation:
One of the main sources of error that occur in a formula of a hydrate lab is that all of the water is not evaporated. We can see at the end of the video that half of the CoCl2 is a light blue colors and the other half is a dark blue color. This indicates that all of the water still has not been evaporated off, resulting in the actual mass of the salt to be greater than the predicted value.
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
Given that
Mass of water = 65.34 g
Amount of heat = mass of water * specific heat (temperature change
)
= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C
= 907.63 J
= 0.908 KJ
And
1 cal = 4.186798 J
907.63 J * 1 cal / 4.186798 J =216.78 cal
Or0.218 kcal
Fossil fuels because easy to transport