Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
Vertical-horizon.
If neither of them are it then it’s columns
In keeping with the general trends, K-Br will have the smallest bond energy. The bond energy refers to the energy that keeps the atoms in a bond together.
<h3>What is bond energy?</h3>
Bond energy is the energy that is required to hold atoms together in a bond. This energy must also be supplied when the atoms are to be separated.
We have the bond energies of each of the bons in the question, we have to note that the the smallest value of bond energy is Na-Br hence in keeping with the general trends, K-Br will have the smallest bond energy.
Learn more about bond energies: brainly.com/question/14842720?
A polar molecule<span> has a net dipole as a result of the opposing charges (i.e. having partial positive and partial negative charges) from </span>polar<span> bonds arranged asymmetrically. Water (H</span>2<span>O) is an example of a </span>polar molecule<span> since it has a slight positive charge on one side and a slight negative charge on the other.</span>
