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Fudgin [204]
3 years ago
5

5. You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second j

ar there is a small amount of water. Using 3 -4 sentences explain how the vapor pressure of water in the first jar compares with the vapor pressure of water in the second jar.
Chemistry
1 answer:
Aneli [31]3 years ago
5 0
The vapor pressure of the two jars are the same.

The volume of water inside the container does not change the vapor pressure.

As long as the liquid remains being water and the temperature does not change the vapor pressure will be the same. The vapor pressure depends only in the nature of the substance and the temperature of the system.
If you want to know more about this, i.e. why, here you have additional explanation:

The vapor pressure is the pressure of the vapor of a substance in equilibrium with the substance in liquid (or solid state) and it is due to the fact that some molecules in the liquid (or solid), those that are close to the surface of liquid in contact with the gas phase and that have enough kinetic energy, evaporate.

At equilibrium the number of molecules passing from the liquid state to the gas state is equal to the number of molecules that pass from the gas state to the liquid state.  If the volume of liquid is increased or decreased, as long as the temperaature of the system remains constant the equilibrium is reached again with the same vapor pressure.


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3 years ago
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A titration of 0.1 M NaOH into 1.2 L of HCl was stopped once the pH reached 7 (at 25C). If 0.4 L of NaOH needed to be added to a
MArishka [77]

Answer:

0.033 M

Explanation:

Let's consider the neutralization reaction between NaOH and HCl.

NaOH + HCl → NaCl + H₂O

0.4 L of 0.1 M NaOH were used. The reacting moles of NaOH are:

0.4 L × 0.1 mol/L = 0.04 mol

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 0.04 moles.

0.04 moles of HCl are in 1.2 L. The molarity of HCl is:

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The atomic number of an atom is always equal to the number of its what ??
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An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
RSB [31]

Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

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Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

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Answer:

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