I think its chlorine Im not positive but im 99.9 precent shure
Answer: 0.050M urea, 0.10M glucose, 0.2M sucrose, pure water
Explanation:
Vapor pressure refers to the ease with which a liquid substance is transformed into vapour. High vapour density implies that the liquid is easily transformed into gas. Pure water is expected to have the lowest vapour density since it is held by strong intermolecular forces in the liquid state. Urea is an organic liquid held by weak Van der Waals forces hence its extremely high vapor pressure.
Answer:
T2 = 2843.1 oK. This is a huge temperature. Check it for errors.
Explanation:
Remark
This is the same question as the other one I've answered. Only the numbers have been altered.
Givens
v1 = 56 mL
P1 = 1 atm
T1 = 273o K
v2 = 162
P2 = 3.6 atm
T2 = ?
Formula
Vi * P1 / T1 = V2 * P2/T2
Solution
Rearrange the formula so T2 is on the left
T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.
T2 = 162 * 3.6* 273 / (56 *1)
T2 = 159213.6/56
T2 = 2843.1
Answer:
Keqq = 310
Note: Some parts of the question were missing. The missing values are used in the explanation below.
Explanation:
<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>
Using the formula ΔG° = -RT㏑Keq
㏑Keq = ΔG°/(-RT)
where ΔG° = ΔH° - TΔS°
㏑Keq = ΔH° - TΔS°/(-RT)
㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)
㏑Keq = -14221.2/-2478.968175
㏑Keq = 5.73674166
Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶
Keq = 310.05
