Answer:
The mol fraction of sucrose in an aqueous solution at 20°C is 0.114
Explanation:
Step 1: Data given
The vapor pressure of water above the solution is 2.0 mmHg
The vapor pressure of pure water at 20 °C is 17.5 mmHg
Step 2: Calculate the mol fraction of sucrose
Psolution = (χsolvent) (P°solvent)
⇒with Psolution = the vapor pressure of the solution is 2.0 mmHg
⇒with χsolvent = the mol fraction of sucrose = TO BE DETERMINED
⇒with P°solvent = the vapor pressure of pure water at 20 °C =17.5 mmHg
2.0 mmHg = X * 17.5 mmHg
X = 2.0 mmHg / 17.5 mmHg
X = 0.114
The mol fraction of sucrose in an aqueous solution at 20°C is 0.114
Answer:
203 grams
Explanation:
The no. of moles of (6.3 x 10²⁴ molecules--Avagadros number) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.
The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.
Oxidation is "Increase in oxidation number" as well as loss of electrons.
A rise in oxidation number results from the loss of negative electrons, whereas a reduction in oxidation number results from the gain of electrons. As a consequence, the oxidized element or ion experiences a rise in oxidation number.
As a result of losing electrons in the process, a reactant oxidizes. When a reactant obtains electrons during a reaction, reduction takes place. This frequently happens when acid and metals react.
Therefore, Oxidation is "Increase in oxidation number" as well as loss of electrons.
Hence, the correct answer will be option (e)
To know more about Oxidation
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Answer:
Your statement is right.
Explanation:
When fuels are burned in vehicle engines, high temperatures are produced. At this high temperature, nitrogen and oxygen from the air combine with each other to produce nitrogen monoxide (NO). When this nitrogen monoxide is released in the air from vehicle exhaust systems, it combines with oxygen present in the air to form nitrogen dioxide (NO2).
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