physical change because even though gas formation was observed, the water was undergoing a state change, which means that its original properties are preserved
Explanation:
The boiling of water is a physical change because the original properties of the water is preserved.
- A physical change is the one that alters the physical properties of matter.
- This kind of change is easily reversible as the water can be cooled back.
- No new kinds of matter is produced in this kind of change .
- The molecules of the matter still retains their property after.
During the boiling of water, intermolecular bonds called hydrogen bonds between the water molecules are broken. This makes the individual molecules free.
Learn more:
Hydrogen bonds brainly.com/question/10602513
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Answer:
1.6 L
Explanation:
Using Charle's law
Given ,
V₁ = 1.5 L
V₂ = ?
T₁ = 12 °C
T₂ = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (12 + 273.15) K = 285.15 K
T₂ = (32 + 273.15) K = 305.15 K
Using above equation as:
New volume = 1.6 L
Answer:
Not exactly But you can take the slope of the curved portion and the slope of the flatline.
It wont do you much good since your working for absorbance but if you ever see something like a temperature change you can use the slope(s) to find freezing points/melting
Explanation:
If you need to submit a slope you could use a best fit which is just point to point or you could break it up like i mentioned
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = ![\frac{ΔvapH}{R}([tex]\frac{1}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7B%CE%94vapH%7D%7BR%7D%28%5Btex%5D%5Cfrac%7B1%7D%7BT1%7D)
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
