Answer:
Answer: What can experiments in a lab tell us about substances on Titan? Experiments in a lab can tell us that the lake did not evaporate in 2007 because the molecular attraction was a lot stronger, then it got weaker overtime.
How does Dr. Hayes' and Dr. Malaska’s research differ? Why are both research projects important? Their research differs because they were both talking about different things, Hayes was talking about how many lakes there were, while Malaska's was doing more hands on stuff like experiments. Both are important because we need to learn how the lakes formed, but we also need to do hands on experiments.
Explanation:
Answer:
1) open system is the system through which both matter and energy and enter as well as leave
ex - open beaker
2) close system is the system through which only energy can leave but matter cannot
ex - pressure cooker.. ( energy leave as air)
3) Valency of an atom is the no. of outer most electron which can participate in a reaction.
4). Heisenberg uncertainty principal states that it is impossible to measure or calculate exactly, both the position and the momentum of an object
5) The reaction in which heat is required to make products are called endothermic reaction. it cools down the surrounding .
Remember significant figures are all non-zero numbers.
Answer:
-1960 kJ.
Step-by-step explanation:
C₄H₄O₄(s) + 3O₂(g) ⟶ 2H₂O((ℓ)) + 4CO₂(g) + Energy
There are three energy flows in this reaction.
From combustion + warm water + warm calorimeter = 0
q₁ + q₂ + q₃ = 0
nΔH + mCwΔT + CcalΔT = 0
<em>Data:
</em>
Mass of fumaric acid = 1.1070 g
Mass of water = 1.093 × 10³ g
Cw = 4.184 J·°C⁻¹g⁻¹
T₁ = 21.10 °C
T₂ = 24.52 °C
Ccal = 891.1 J·°C⁻¹
Calculations:
(a) <em>q₁
</em>
n = 1.1070 g × (1 mol/116.07 g)
n = 0.009 537 mol
q₁ = 0.009 537ΔH J
(b) <em>q₂
</em>
ΔT = 24.52 – 21.10
ΔT = 3.42°C
q₂ = 1093 × 4.184 × 3.42
q₂ = 15 640 J
(c) <em>q₃
</em>
q₃ = 891.1 × 3.42
q₃ =3048 J
(d) <em>ΔH</em>
0.009 537ΔH + 15 640 + 3048 = 0
0.009 537ΔH + 18 688 = 0
0.009 537ΔH = -18 688
ΔH = -18 688/0.009 537
ΔH = -1 959 413 J/mol
ΔH = -1960 kJ/mol
This is quite different from the actual value of -1334.70 kJ·mol⁻¹