You would observe, trying to identify if there's a minimum difference in each beaker. Like, colour, or using an <em>bagueta </em>(idk how to say it in english), churning it or something like that.
<span> It has a very HIGH melting point that is characteristic of </span>ionic compounds<span>. Think of Table salt as an example. It Melts at very high </span>temperatures<span>. </span>Molecular compounds<span> are easy to melt.3
3 </span><span>https://quizlet.com/61014423/chemistry-81-82-flash-cards/</span>
I personally try to balance the equation without looking at the given coefficients in the unbalanced reaction and this is my result:
2 C2H14+11 O2--> 4 CO2+14 H2O
I think none of the given answers are correct, and the answer is obviously 11.
Hope this will help :))
Answer:
56.6g
Explanation:
Given that :
Mass of water, m1= 300.0 g
Temperature of water, T1= 22°C
Specific heat capacity of water, C= 4.184 J/g°C
Mass of polystyrene, m2=?
Temperature of polystyrene = 94.9
Specific heat capacity of polystyrene, c2= 1.88 J/g°C
Final temperature = 27.7 oC
Heat lost by polystyrene = Heat gained by water
mc(dT) = mc(dT)
m2 * 1.88 * (94.9 - 27.7) = 300 * 4.184 * (27.7 - 22)
m2 * 1.88*67.2 = 300*4.184*5.7
126.336 * m2 = 7154.64
m2 = 7154.64 / 126.336
m2 = 56.631838
Hence, mass of polystyrene = 56.6g
The heat of reaction : 50.6 kJ
<h3>Further explanation</h3>
Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways
Reaction
N₂(g) + 2H₂(g) ⇒N₂H₄(l)
thermochemical data:
1. N₂H₄(l)+O₂(g)⇒N₂(g)+2H₂O(l) ΔH=-622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ
We arrange the position of the elements / compounds so that they correspond to the main reaction, and the enthalpy sign will also change
1. N₂(g)+H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2. H₂(g)+1/2O₂(g)⇒H₂O(l) ΔH=-285.8 kJ x 2 ⇒
2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
Add reaction 1 and reaction 2, and remove the same compound from different sides
1. N₂(g)+2H₂O(l) ⇒ N₂H₄(l)+O₂(g) ΔH=+622.2 kJ
2.2H₂(g)+O₂(g)⇒2H₂O(l) ΔH=-571.6 kJ
-------------------------------------------------------------------- +
N₂(g) + 2H₂(g) ⇒N₂H₄(l) ΔH=50.6 kJ